OP Malhotra Class 11 Maths Solutions Chapter 8 Mathematical Induction Exercise 8 (A)

Use the principle of mathematical induction to prove the following statements for all n \( \in \) N.

Question 1. Prove that \( 1 + 2 + 3 + \ldots + n = \frac{1}{2} n(n + 1) \).
Answer: Let the given statement be \( P(n) : 1 + 2 + 3 + \ldots + n = \frac{1}{2} n(n + 1) \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the Left Hand Side (LHS) is \( 1 \).
The Right Hand Side (RHS) is \( \frac{1}{2} \times 1 \times (1 + 1) = \frac{1}{2} \times 1 \times 2 = 1 \).
Since LHS = RHS, \( P(1) \) is true. This shows the formula works for the first natural number.
Step 2: Inductive Hypothesis
Assume that \( P(k) \) is true for some natural number \( k \).
So, \( 1 + 2 + 3 + \ldots + k = \frac{1}{2} k(k + 1) \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(k+1) \) is true.
Consider the LHS for \( n = k+1 \):
\( 1 + 2 + 3 + \ldots + k + (k + 1) \)
Using Equation (1) for the sum up to \( k \):
\( = \frac{1}{2} k(k + 1) + (k + 1) \)
Take \( (k + 1) \) as a common factor:
\( = (k + 1) \left( \frac{k}{2} + 1 \right) \)
\( = (k + 1) \left( \frac{k + 2}{2} \right) \)
\( = \frac{1}{2} (k + 1) (k + 2) \)
We can write \( k + 2 \) as \( (k + 1 + 1) \).
So, \( = \frac{1}{2} (k + 1) ( (k + 1) + 1 ) \). This is exactly the RHS of \( P(n) \) with \( n \) replaced by \( k+1 \).
Therefore, \( P(k+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(k+1) \) is true whenever \( P(k) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This means the formula works for all counting numbers.
In simple words: We showed the formula works for the first number (1). Then, we assumed it works for any number 'k' and used that to prove it also works for 'k+1'. Because of this chain reaction, the formula is true for all natural numbers.

🎯 Exam Tip: When using mathematical induction, always clearly state the base case, inductive hypothesis, and inductive step. Factoring out common terms like \( (k+1) \) is often a crucial step in simplifying the expression to match the form for \( n=k+1 \).

Question 2. Prove that \( 2 + 4 + 6 + \ldots + 2n = n (n + 1) \).
Answer: Let the given statement be \( P(n) : 2 + 4 + 6 + \ldots + 2n = n (n + 1) \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( 2 \).
The RHS is \( 1 (1 + 1) = 1 \times 2 = 2 \).
Since LHS = RHS, \( P(1) \) is true. This means the formula works for the first natural number.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( 2 + 4 + 6 + \ldots + 2m = m (m + 1) \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( 2 + 4 + 6 + \ldots + 2(m+1) = (m+1)(m+2) \).
Consider the LHS for \( n = m+1 \):
\( 2 + 4 + 6 + \ldots + 2m + 2(m + 1) \)
Using Equation (1) for the sum up to \( 2m \):
\( = m (m + 1) + 2(m + 1) \)
Take \( (m + 1) \) as a common factor:
\( = (m + 1) (m + 2) \)
This is exactly the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This process ensures the statement holds for all subsequent numbers.
In simple words: First, we prove the statement for the smallest number (1). Then, we assume it's true for any number 'm' and show that it must then also be true for 'm+1'. This 'domino effect' proves the statement for all natural numbers.

🎯 Exam Tip: For series questions, carefully identify the 'k+1' term and separate it from the sum up to 'k' to apply the inductive hypothesis effectively.

Question 3. Prove that \( 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{1}{6} n(n + 1)(2n + 1) \).
Answer: Let the given statement be \( P(n) : 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{1}{6} n(n + 1)(2n + 1) \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( 1^2 = 1 \).
The RHS is \( \frac{1}{6} \times 1 \times (1 + 1) \times (2 \times 1 + 1) = \frac{1}{6} \times 1 \times 2 \times 3 = \frac{6}{6} = 1 \).
Since LHS = RHS, \( P(1) \) is true. This confirms the formula works for the first case.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( 1^2 + 2^2 + 3^2 + \ldots + m^2 = \frac{1}{6} m(m + 1)(2m + 1) \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( 1^2 + 2^2 + \ldots + (m+1)^2 = \frac{1}{6} (m+1)((m+1)+1)(2(m+1)+1) \).
Which simplifies to \( \frac{1}{6} (m+1)(m+2)(2m+3) \).
Consider the LHS for \( n = m+1 \):
\( 1^2 + 2^2 + \ldots + m^2 + (m + 1)^2 \)
Using Equation (1) for the sum up to \( m^2 \):
\( = \frac{1}{6} m(m + 1)(2m + 1) + (m + 1)^2 \)
Take \( (m + 1) \) as a common factor:
\( = (m + 1) \left[ \frac{m(2m + 1)}{6} + (m + 1) \right] \)
\( = (m + 1) \left[ \frac{2m^2 + m + 6(m + 1)}{6} \right] \)
\( = (m + 1) \left[ \frac{2m^2 + m + 6m + 6}{6} \right] \)
\( = (m + 1) \left[ \frac{2m^2 + 7m + 6}{6} \right] \)
Now, factor the quadratic term \( (2m^2 + 7m + 6) \). We look for two numbers that multiply to \( 2 \times 6 = 12 \) and add to \( 7 \), which are 3 and 4.
\( 2m^2 + 7m + 6 = 2m^2 + 4m + 3m + 6 \)
\( = 2m(m + 2) + 3(m + 2) \)
\( = (2m + 3)(m + 2) \)
So, the expression becomes:
\( = (m + 1) \left[ \frac{(2m + 3)(m + 2)}{6} \right] \)
\( = \frac{1}{6} (m + 1)(m + 2)(2m + 3) \)
Rewrite \( m+2 \) as \( (m+1)+1 \) and \( 2m+3 \) as \( 2(m+1)+1 \):
\( = \frac{1}{6} (m + 1) ( (m + 1) + 1 ) ( 2(m + 1) + 1 ) \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This method confirms the sum of squares formula for any natural number.
In simple words: We first showed the formula is true for the number 1. Then, we assumed it works for a number 'm' and used that to prove it also works for 'm+1'. This 'step-by-step' logic means it holds true for all natural numbers.

🎯 Exam Tip: Remember to factorize the quadratic expression correctly in the inductive step. A common mistake is to miss the factors, which prevents matching the form for \( n=k+1 \).

Question 4. Prove that \( 1^2 + 3^2 + 5^2 + \ldots + (2n - 1)^2 = \frac{1}{3} n(2n - 1)(2n + 1) \).
Answer: Let the given statement be \( P(n) : 1^2 + 3^2 + 5^2 + \ldots + (2n - 1)^2 = \frac{1}{3} n(2n - 1)(2n + 1) \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( 1^2 = 1 \).
The RHS is \( \frac{1}{3} \times 1 \times (2 \times 1 - 1) \times (2 \times 1 + 1) = \frac{1}{3} \times 1 \times 1 \times 3 = 1 \).
Since LHS = RHS, \( P(1) \) is true. This verifies the formula for the initial value.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( 1^2 + 3^2 + 5^2 + \ldots + (2m - 1)^2 = \frac{1}{3} m(2m - 1)(2m + 1) \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( 1^2 + 3^2 + \ldots + (2(m+1)-1)^2 = \frac{1}{3} (m+1)(2(m+1)-1)(2(m+1)+1) \).
Which simplifies to \( \frac{1}{3} (m+1)(2m+1)(2m+3) \).
Consider the LHS for \( n = m+1 \):
\( 1^2 + 3^2 + 5^2 + \ldots + (2m - 1)^2 + (2(m + 1) - 1)^2 \)
\( = 1^2 + 3^2 + 5^2 + \ldots + (2m - 1)^2 + (2m + 1)^2 \)
Using Equation (1) for the sum up to \( (2m - 1)^2 \):
\( = \frac{1}{3} m(2m - 1)(2m + 1) + (2m + 1)^2 \)
Take \( (2m + 1) \) as a common factor:
\( = (2m + 1) \left[ \frac{m(2m - 1)}{3} + (2m + 1) \right] \)
\( = (2m + 1) \left[ \frac{m(2m - 1) + 3(2m + 1)}{3} \right] \)
\( = (2m + 1) \left[ \frac{2m^2 - m + 6m + 3}{3} \right] \)
\( = (2m + 1) \left[ \frac{2m^2 + 5m + 3}{3} \right] \)
Now, factor the quadratic term \( (2m^2 + 5m + 3) \). We look for two numbers that multiply to \( 2 \times 3 = 6 \) and add to \( 5 \), which are 2 and 3.
\( 2m^2 + 5m + 3 = 2m^2 + 2m + 3m + 3 \)
\( = 2m(m + 1) + 3(m + 1) \)
\( = (2m + 3)(m + 1) \)
So, the expression becomes:
\( = (2m + 1) \left[ \frac{(m + 1)(2m + 3)}{3} \right] \)
\( = \frac{1}{3} (m + 1)(2m + 1)(2m + 3) \)
Rewrite \( 2m+1 \) as \( 2(m+1)-1 \) and \( 2m+3 \) as \( 2(m+1)+1 \):
\( = \frac{1}{3} (m + 1) ( 2(m + 1) - 1 ) ( 2(m + 1) + 1 ) \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This process of building from one step to the next ensures the formula holds for all applicable numbers.
In simple words: We first showed the formula is correct for the smallest odd square (1 squared). Then, we assumed it works for any 'm' terms and proved it works for 'm+1' terms. This means the pattern is always true for any natural number 'n'.

🎯 Exam Tip: When dealing with sums of odd numbers, the term is \( (2n-1)^2 \). Ensure the \( k+1 \) term is correctly written as \( (2(k+1)-1)^2 = (2k+1)^2 \).

Question 5. Prove that \( 2 + 2^2 + 2^3 + \ldots + 2^n = 2 (2^n - 1) \).
Answer: Let the given statement be \( P(n) : 2 + 2^2 + 2^3 + \ldots + 2^n = 2 (2^n - 1) \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( 2 \).
The RHS is \( 2 (2^1 - 1) = 2 (2 - 1) = 2 \times 1 = 2 \).
Since LHS = RHS, \( P(1) \) is true. This shows the formula works for the first term.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( 2 + 2^2 + 2^3 + \ldots + 2^m = 2 (2^m - 1) \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( 2 + 2^2 + \ldots + 2^{m+1} = 2(2^{m+1} - 1) \).
Consider the LHS for \( n = m+1 \):
\( (2 + 2^2 + 2^3 + \ldots + 2^m) + 2^{m+1} \)
Using Equation (1) for the sum up to \( 2^m \):
\( = 2(2^m - 1) + 2^{m+1} \)
\( = 2 \times 2^m - 2 + 2^{m+1} \)
\( = 2^{m+1} - 2 + 2^{m+1} \)
Combine the \( 2^{m+1} \) terms:
\( = 2 \times 2^{m+1} - 2 \)
Take \( 2 \) as a common factor:
\( = 2 (2^{m+1} - 1) \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This confirms the sum of powers of 2 for any natural number.
In simple words: We first checked if the rule works for the number one. Then, we supposed it works for any number 'm' and used that idea to show it also works for the next number 'm+1'. This means the rule is true for all whole numbers.

🎯 Exam Tip: When working with powers in induction, remember exponent rules like \( a^x \times a^y = a^{x+y} \) and \( a^x + a^x = 2 \times a^x \). These are vital for simplifying expressions.

Question 6. Prove that \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n(n + 1) = \frac{n(n+1)(n+2)}{3} \).
Answer: Let the given statement be \( P(n) : 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \ldots + n(n + 1) = \frac{n(n+1)(n+2)}{3} \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( 1 \cdot 2 = 2 \).
The RHS is \( \frac{1(1+1)(1+2)}{3} = \frac{1 \times 2 \times 3}{3} = 2 \).
Since LHS = RHS, \( P(1) \) is true. This proves the formula for the first term.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( 1 \cdot 2 + 2 \cdot 3 + \ldots + m(m + 1) = \frac{m(m+1)(m+2)}{3} \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( 1 \cdot 2 + 2 \cdot 3 + \ldots + (m+1)((m+1)+1) = \frac{(m+1)((m+1)+1)((m+1)+2)}{3} \).
Which simplifies to \( \frac{(m+1)(m+2)(m+3)}{3} \).
Consider the LHS for \( n = m+1 \):
\( (1 \cdot 2 + 2 \cdot 3 + \ldots + m(m + 1)) + (m + 1)((m + 1) + 1) \)
\( = (1 \cdot 2 + 2 \cdot 3 + \ldots + m(m + 1)) + (m + 1)(m + 2) \)
Using Equation (1) for the sum up to \( m(m+1) \):
\( = \frac{m(m+1)(m+2)}{3} + (m + 1)(m + 2) \)
Take \( (m + 1)(m + 2) \) as a common factor:
\( = (m + 1)(m + 2) \left( \frac{m}{3} + 1 \right) \)
\( = (m + 1)(m + 2) \left( \frac{m + 3}{3} \right) \)
\( = \frac{(m + 1)(m + 2)(m + 3)}{3} \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This confirms the pattern of these products for all counting numbers.
In simple words: We first showed the formula is true for the smallest product term. Then, we assumed it's true for 'm' terms and proved it also works for 'm+1' terms. This step-by-step proof confirms the formula for all natural numbers.

🎯 Exam Tip: Recognizing common factors like \( (m+1)(m+2) \) in the inductive step simplifies the algebra significantly and is key to reaching the desired form.

Question 7. Prove that \( 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \ldots + n \cdot 2^n = (n - 1) 2^{n+1} + 2 \).
Answer: Let the given statement be \( P(n) : 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + \ldots + n \cdot 2^n = (n - 1) 2^{n+1} + 2 \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( 1 \cdot 2 = 2 \).
The RHS is \( (1 - 1) 2^{1+1} + 2 = 0 \cdot 2^2 + 2 = 0 + 2 = 2 \).
Since LHS = RHS, \( P(1) \) is true. This confirms the formula for the first term.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( 1 \cdot 2 + 2 \cdot 2^2 + \ldots + m \cdot 2^m = (m - 1) 2^{m+1} + 2 \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( 1 \cdot 2 + 2 \cdot 2^2 + \ldots + (m+1) 2^{m+1} = ((m+1) - 1) 2^{((m+1)+1)} + 2 \).
Which simplifies to \( m \cdot 2^{m+2} + 2 \).
Consider the LHS for \( n = m+1 \):
\( (1 \cdot 2 + 2 \cdot 2^2 + \ldots + m \cdot 2^m) + (m + 1) 2^{m+1} \)
Using Equation (1) for the sum up to \( m \cdot 2^m \):
\( = (m - 1) 2^{m+1} + 2 + (m + 1) 2^{m+1} \)
Group the terms with \( 2^{m+1} \):
\( = ( (m - 1) + (m + 1) ) 2^{m+1} + 2 \)
\( = ( m - 1 + m + 1 ) 2^{m+1} + 2 \)
\( = (2m) 2^{m+1} + 2 \)
\( = m \cdot (2 \cdot 2^{m+1}) + 2 \)
\( = m \cdot 2^{m+2} + 2 \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This shows that the formula holds for every natural number through a step-by-step verification.
In simple words: We first proved the formula works for 'n=1'. Then, we assumed it works for 'n=m' and used this assumption to show it also works for 'n=m+1'. This method guarantees the formula is true for all natural numbers.

🎯 Exam Tip: Be careful with algebra involving exponents. Combine like terms by factoring out powers, e.g., \( (m-1)2^{m+1} + (m+1)2^{m+1} \). Ensure to distribute and re-group terms correctly.

Question 8. Prove that \( 3 \cdot 2^2 + 3^2 \cdot 2^3 + 3^3 \cdot 2^4 + \ldots + 3^n \cdot 2^{n+1} = \frac{12}{5} (6^n - 1) \).
Answer: Let the given statement be \( P(n) : 3 \cdot 2^2 + 3^2 \cdot 2^3 + 3^3 \cdot 2^4 + \ldots + 3^n \cdot 2^{n+1} = \frac{12}{5} (6^n - 1) \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( 3 \cdot 2^2 = 3 \cdot 4 = 12 \).
The RHS is \( \frac{12}{5} (6^1 - 1) = \frac{12}{5} (6 - 1) = \frac{12}{5} \times 5 = 12 \).
Since LHS = RHS, \( P(1) \) is true. This shows the formula works for the first term.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( 3 \cdot 2^2 + 3^2 \cdot 2^3 + \ldots + 3^m \cdot 2^{m+1} = \frac{12}{5} (6^m - 1) \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( 3 \cdot 2^2 + \ldots + 3^{m+1} \cdot 2^{m+2} = \frac{12}{5} (6^{m+1} - 1) \).
Consider the LHS for \( n = m+1 \):
\( (3 \cdot 2^2 + 3^2 \cdot 2^3 + \ldots + 3^m \cdot 2^{m+1}) + 3^{m+1} \cdot 2^{m+2} \)
Using Equation (1) for the sum up to \( 3^m \cdot 2^{m+1} \):
\( = \frac{12}{5} (6^m - 1) + 3^{m+1} \cdot 2^{m+2} \)
Recall that \( 3^{m+1} \cdot 2^{m+2} = (3 \cdot 2)^{m+1} \cdot 2 = 6^{m+1} \cdot 2 \). Also, we can write \( 3^{m+1} \cdot 2^{m+2} = 3^m \cdot 3 \cdot 2^m \cdot 2^2 = (3^m \cdot 2^m) \cdot (3 \cdot 4) = 6^m \cdot 12 \).
\( = \frac{12}{5} (6^m - 1) + 12 \cdot 6^m \)
\( = \frac{12}{5} 6^m - \frac{12}{5} + 12 \cdot 6^m \)
Combine the \( 6^m \) terms:
\( = 6^m \left( \frac{12}{5} + 12 \right) - \frac{12}{5} \)
\( = 6^m \left( \frac{12 + 60}{5} \right) - \frac{12}{5} \)
\( = 6^m \left( \frac{72}{5} \right) - \frac{12}{5} \)
Now, we want to reach the form \( \frac{12}{5} (6^{m+1} - 1) \).
\( \frac{72}{5} 6^m - \frac{12}{5} = \frac{12}{5} (6 \cdot 6^m - 1) \)
\( = \frac{12}{5} (6^{m+1} - 1) \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This demonstrates the formula's consistency across all natural numbers.
In simple words: We first checked if the formula is true for the number one. Then, we assumed it is true for any number 'm' and used that to show it must also be true for 'm+1'. This 'domino effect' means the formula works for every natural number.

🎯 Exam Tip: When the terms involve products of powers (like \( 3^n \cdot 2^{n+1} \)), it's helpful to express them in a common base (like \( 6^n \)) to simplify the inductive step. Be careful with calculations of the coefficients.

Question 9. Prove that \( 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \ldots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3} \).
Answer: Let the given statement be \( P(n) : 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \ldots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3} \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( 1 \cdot 3 = 3 \).
The RHS is \( \frac{1(4 \cdot 1^2 + 6 \cdot 1 - 1)}{3} = \frac{1(4 + 6 - 1)}{3} = \frac{9}{3} = 3 \).
Since LHS = RHS, \( P(1) \) is true. This confirms the formula for the first term.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( 1 \cdot 3 + 3 \cdot 5 + \ldots + (2m - 1)(2m + 1) = \frac{m(4m^2 + 6m - 1)}{3} \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( P(m+1) = \frac{(m+1)(4(m+1)^2 + 6(m+1) - 1)}{3} \).
Consider the LHS for \( n = m+1 \):
\( (1 \cdot 3 + 3 \cdot 5 + \ldots + (2m - 1)(2m + 1)) + (2(m+1) - 1)(2(m+1) + 1) \)
\( = (1 \cdot 3 + 3 \cdot 5 + \ldots + (2m - 1)(2m + 1)) + (2m + 1)(2m + 3) \)
Using Equation (1) for the sum up to \( (2m - 1)(2m + 1) \):
\( = \frac{m(4m^2 + 6m - 1)}{3} + (2m + 1)(2m + 3) \)
To combine, find a common denominator:
\( = \frac{m(4m^2 + 6m - 1) + 3(2m + 1)(2m + 3)}{3} \)
Expand the terms:
\( = \frac{4m^3 + 6m^2 - m + 3(4m^2 + 6m + 2m + 3)}{3} \)
\( = \frac{4m^3 + 6m^2 - m + 3(4m^2 + 8m + 3)}{3} \)
\( = \frac{4m^3 + 6m^2 - m + 12m^2 + 24m + 9}{3} \)
\( = \frac{4m^3 + 18m^2 + 23m + 9}{3} \)
Now we need to show that this is equal to \( \frac{(m+1)(4(m+1)^2 + 6(m+1) - 1)}{3} \).
Let's expand the target RHS expression:
\( (m+1)(4(m^2+2m+1) + 6m+6 - 1) \)
\( = (m+1)(4m^2+8m+4 + 6m+5) \)
\( = (m+1)(4m^2+14m+9) \)
\( = 4m^3 + 14m^2 + 9m + 4m^2 + 14m + 9 \)
\( = 4m^3 + 18m^2 + 23m + 9 \)
Since our derived expression matches the target expression, \( P(m+1) \) is true.
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This demonstrates the reliability of the formula across all natural numbers.
In simple words: We first showed the formula is correct for 'n=1'. Then, we assumed it works for 'n=m' and used that to show it also works for 'n=m+1'. This step-by-step logic proves the formula is true for all natural numbers.

🎯 Exam Tip: When the resulting expression is complex, it's often easier to expand the target \( P(k+1) \) formula and compare it to your simplified LHS rather than trying to factorize a high-degree polynomial.

Question 10. Prove the following by the method of mathematical induction:
(i) \( a + (a + d) + (a + 2d) + \ldots + [a + (n - 1)d] = \frac{n}{2} [2a + (n - 1)d] \)
(ii) \( a + ar + ar^2 + \ldots + ar^{n-1} = \frac{a(1 - r^n)}{1 - r} \), where \( r \neq 1 \)
Answer:
(i) Let the given statement be \( P(n) : a + (a + d) + (a + 2d) + \ldots + [a + (n - 1)d] = \frac{n}{2} [2a + (n - 1)d] \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( a \).
The RHS is \( \frac{1}{2} [2a + (1 - 1)d] = \frac{1}{2} [2a + 0] = a \).
Since LHS = RHS, \( P(1) \) is true. This means the formula works for the first term of an arithmetic progression.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( a + (a + d) + \ldots + [a + (m - 1)d] = \frac{m}{2} [2a + (m - 1)d] \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( P(m+1) = \frac{(m+1)}{2} [2a + ((m+1) - 1)d] = \frac{(m+1)}{2} [2a + md] \).
Consider the LHS for \( n = m+1 \):
\( (a + (a + d) + \ldots + [a + (m - 1)d]) + [a + ((m+1) - 1)d] \)
\( = (a + (a + d) + \ldots + [a + (m - 1)d]) + (a + md) \)
Using Equation (1) for the sum up to \( [a + (m - 1)d] \):
\( = \frac{m}{2} [2a + (m - 1)d] + (a + md) \)
\( = \frac{m(2a + md - d)}{2} + (a + md) \)
\( = \frac{2am + m^2d - md + 2a + 2md}{2} \)
\( = \frac{2am + m^2d + md + 2a}{2} \)
Factor out \( (m+1) \):
\( = \frac{2a(m+1) + md(m+1)}{2} \)
\( = \frac{(m+1)(2a + md)}{2} \)
This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion for (i)
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This confirms the formula for the sum of an arithmetic progression.

(ii) Let the given statement be \( P(n) : a + ar + ar^2 + \ldots + ar^{n-1} = \frac{a(1 - r^n)}{1 - r} \), where \( r \neq 1 \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( a \).
The RHS is \( \frac{a(1 - r^1)}{1 - r} = a \).
Since LHS = RHS, \( P(1) \) is true. This verifies the formula for the first term of a geometric progression.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( a + ar + ar^2 + \ldots + ar^{m-1} = \frac{a(1 - r^m)}{1 - r} \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( P(m+1) = \frac{a(1 - r^{m+1})}{1 - r} \).
Consider the LHS for \( n = m+1 \):
\( (a + ar + ar^2 + \ldots + ar^{m-1}) + ar^{(m+1)-1} \)
\( = (a + ar + ar^2 + \ldots + ar^{m-1}) + ar^m \)
Using Equation (1) for the sum up to \( ar^{m-1} \):
\( = \frac{a(1 - r^m)}{1 - r} + ar^m \)
To combine, find a common denominator:
\( = \frac{a(1 - r^m) + ar^m(1 - r)}{1 - r} \)
\( = \frac{a - ar^m + ar^m - ar^{m+1}}{1 - r} \)
The \( -ar^m \) and \( +ar^m \) terms cancel out:
\( = \frac{a - ar^{m+1}}{1 - r} \)
Factor out \( a \):
\( = \frac{a(1 - r^{m+1})}{1 - r} \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion for (ii)
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This method universally confirms the sum of a geometric progression.
In simple words: For both arithmetic and geometric sequences, we first showed the formula works for the first term. Then, we assumed it works for 'm' terms and proved it also works for 'm+1' terms. This logical step-by-step process confirms that both formulas are true for any number of terms.

🎯 Exam Tip: When proving formulas for arithmetic progressions (AP) and geometric progressions (GP) using induction, ensure you correctly identify the common difference 'd' or common ratio 'r' and substitute the \( k+1 \) term accurately. Algebraic simplification is crucial here.

Question 11. Prove that \( 5 + 15 + 45 + \ldots + 5 \cdot 3^{n-1} = \frac{5}{2} (3^n - 1) \).
Answer: Let the given statement be \( P(n) : 5 + 15 + 45 + \ldots + 5 \cdot 3^{n-1} = \frac{5}{2} (3^n - 1) \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( 5 \).
The RHS is \( \frac{5}{2} (3^1 - 1) = \frac{5}{2} (3 - 1) = \frac{5}{2} \times 2 = 5 \).
Since LHS = RHS, \( P(1) \) is true. This shows the formula works for the first term.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( 5 + 15 + 45 + \ldots + 5 \cdot 3^{m-1} = \frac{5}{2} (3^m - 1) \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( P(m+1) = \frac{5}{2} (3^{m+1} - 1) \).
Consider the LHS for \( n = m+1 \):
\( (5 + 15 + 45 + \ldots + 5 \cdot 3^{m-1}) + 5 \cdot 3^{(m+1)-1} \)
\( = (5 + 15 + 45 + \ldots + 5 \cdot 3^{m-1}) + 5 \cdot 3^m \)
Using Equation (1) for the sum up to \( 5 \cdot 3^{m-1} \):
\( = \frac{5}{2} (3^m - 1) + 5 \cdot 3^m \)
To combine, find a common denominator:
\( = \frac{5(3^m - 1) + 2 \cdot 5 \cdot 3^m}{2} \)
\( = \frac{5 \cdot 3^m - 5 + 10 \cdot 3^m}{2} \)
Combine the \( 3^m \) terms:
\( = \frac{(5 + 10) \cdot 3^m - 5}{2} \)
\( = \frac{15 \cdot 3^m - 5}{2} \)
Factor out \( 5 \):
\( = \frac{5(3 \cdot 3^m - 1)}{2} \)
\( = \frac{5(3^{m+1} - 1)}{2} \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This demonstrates the formula's consistency for all such sums.
In simple words: We first checked if the rule holds for the starting number (n=1). Then, we assumed it works for any number 'm' and used that to show it also works for the next number 'm+1'. This step-by-step proof means the rule is true for all natural numbers.

🎯 Exam Tip: When simplifying expressions involving powers, remember to combine terms like \( a \cdot b^m + c \cdot b^m = (a+c) b^m \) and \( x \cdot b^m = b^{m+ \text{log}_b x} \). This helps in reaching the final form.

Question 12. Prove that \( \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \ldots + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)} \).
Answer: Let the given statement be \( P(n) : \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} + \ldots + \frac{1}{(2n + 1)(2n + 3)} = \frac{n}{3(2n + 3)} \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( \frac{1}{3 \cdot 5} = \frac{1}{15} \).
The RHS is \( \frac{1}{3(2 \cdot 1 + 3)} = \frac{1}{3(5)} = \frac{1}{15} \).
Since LHS = RHS, \( P(1) \) is true. This shows the formula works for the first term.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2m + 1)(2m + 3)} = \frac{m}{3(2m + 3)} \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( P(m+1) = \frac{(m+1)}{3(2(m+1) + 3)} = \frac{(m+1)}{3(2m + 5)} \).
Consider the LHS for \( n = m+1 \):
\( \left( \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2m + 1)(2m + 3)} \right) + \frac{1}{(2(m+1) + 1)(2(m+1) + 3)} \)
\( = \left( \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2m + 1)(2m + 3)} \right) + \frac{1}{(2m + 3)(2m + 5)} \)
Using Equation (1) for the sum up to \( \frac{1}{(2m + 1)(2m + 3)} \):
\( = \frac{m}{3(2m + 3)} + \frac{1}{(2m + 3)(2m + 5)} \)
To combine, find a common denominator, which is \( 3(2m+3)(2m+5) \):
\( = \frac{m(2m + 5) + 3}{3(2m + 3)(2m + 5)} \)
\( = \frac{2m^2 + 5m + 3}{3(2m + 3)(2m + 5)} \)
Now, factor the quadratic term in the numerator \( (2m^2 + 5m + 3) \). We look for two numbers that multiply to \( 2 \times 3 = 6 \) and add to \( 5 \), which are 2 and 3.
\( 2m^2 + 5m + 3 = 2m^2 + 2m + 3m + 3 \)
\( = 2m(m + 1) + 3(m + 1) \)
\( = (2m + 3)(m + 1) \)
So, the expression becomes:
\( = \frac{(m + 1)(2m + 3)}{3(2m + 3)(2m + 5)} \)
Cancel out \( (2m + 3) \) from the numerator and denominator:
\( = \frac{m + 1}{3(2m + 5)} \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This demonstrates the formula's accuracy for all natural numbers.
In simple words: We showed the formula works for the first term. Then, we assumed it works for any 'm' terms and used that to prove it also works for 'm+1' terms. This means the formula is always true for any natural number.

🎯 Exam Tip: When dealing with fractional series, partial fraction decomposition is often the basis of the formula. In the inductive step, be careful with finding a common denominator and factoring the numerator to cancel terms, which simplifies the expression to the desired form.

Question 13. Prove that \( \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1} \).
Answer: Let the given statement be \( P(n) : \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots + \frac{1}{(3n - 2)(3n + 1)} = \frac{n}{3n + 1} \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( \frac{1}{1 \cdot 4} = \frac{1}{4} \).
The RHS is \( \frac{1}{3 \cdot 1 + 1} = \frac{1}{4} \).
Since LHS = RHS, \( P(1) \) is true. This confirms the formula for the first term.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \ldots + \frac{1}{(3m - 2)(3m + 1)} = \frac{m}{3m + 1} \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( P(m+1) = \frac{(m+1)}{3(m+1) + 1} = \frac{m+1}{3m + 4} \).
Consider the LHS for \( n = m+1 \):
\( \left( \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \ldots + \frac{1}{(3m - 2)(3m + 1)} \right) + \frac{1}{(3(m+1) - 2)(3(m+1) + 1)} \)
\( = \left( \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \ldots + \frac{1}{(3m - 2)(3m + 1)} \right) + \frac{1}{(3m + 3 - 2)(3m + 3 + 1)} \)
\( = \left( \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \ldots + \frac{1}{(3m - 2)(3m + 1)} \right) + \frac{1}{(3m + 1)(3m + 4)} \)
Using Equation (1) for the sum up to \( \frac{1}{(3m - 2)(3m + 1)} \):
\( = \frac{m}{3m + 1} + \frac{1}{(3m + 1)(3m + 4)} \)
To combine, find a common denominator, which is \( (3m+1)(3m+4) \):
\( = \frac{m(3m + 4) + 1}{(3m + 1)(3m + 4)} \)
\( = \frac{3m^2 + 4m + 1}{(3m + 1)(3m + 4)} \)
Now, factor the quadratic term in the numerator \( (3m^2 + 4m + 1) \). We look for two numbers that multiply to \( 3 \times 1 = 3 \) and add to \( 4 \), which are 1 and 3.
\( 3m^2 + 4m + 1 = 3m^2 + 3m + m + 1 \)
\( = 3m(m + 1) + 1(m + 1) \)
\( = (3m + 1)(m + 1) \)
So, the expression becomes:
\( = \frac{(m + 1)(3m + 1)}{(3m + 1)(3m + 4)} \)
Cancel out \( (3m + 1) \) from the numerator and denominator:
\( = \frac{m + 1}{3m + 4} \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This process ensures the formula holds for all possible natural numbers.
In simple words: We first checked if the formula is true for the number one. Then, we assumed it works for any 'm' and used that to show it also works for 'm+1'. This means the formula is true for all natural numbers.

🎯 Exam Tip: Similar to other fractional series, a key step is factoring the numerator after combining terms to allow for cancellation with the denominator. Careful algebraic manipulation is crucial.

Question 14. Prove that \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^n} = 1 - \frac{1}{2^n} \).
Answer: Let the given statement be \( P(n) : \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^n} = 1 - \frac{1}{2^n} \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( \frac{1}{2} \).
The RHS is \( 1 - \frac{1}{2^1} = 1 - \frac{1}{2} = \frac{1}{2} \).
Since LHS = RHS, \( P(1) \) is true. This shows the formula works for the first term.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^m} = 1 - \frac{1}{2^m} \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( P(m+1) = 1 - \frac{1}{2^{m+1}} \).
Consider the LHS for \( n = m+1 \):
\( \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^m} \right) + \frac{1}{2^{m+1}} \)
Using Equation (1) for the sum up to \( \frac{1}{2^m} \):
\( = \left( 1 - \frac{1}{2^m} \right) + \frac{1}{2^{m+1}} \)
Rewrite \( \frac{1}{2^m} \) as \( \frac{2}{2 \cdot 2^m} = \frac{2}{2^{m+1}} \).
\( = 1 - \frac{2}{2^{m+1}} + \frac{1}{2^{m+1}} \)
\( = 1 - \frac{2 - 1}{2^{m+1}} \)
\( = 1 - \frac{1}{2^{m+1}} \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This demonstrates the sum of this geometric series works for all natural numbers.
In simple words: We proved the formula for the first term. Then, by assuming it's true for 'm' terms, we showed it must also be true for 'm+1' terms. This confirms the formula works for all natural numbers.

🎯 Exam Tip: Manipulating fractions with exponents often requires finding a common denominator and carefully combining terms. Remember that \( \frac{1}{2^m} = \frac{2}{2^{m+1}} \) to combine fractions easily.

Question 15. Prove that \( 1 + 5 + 12 + 22 + 35 + \ldots + \frac{n(3n - 1)}{2} = \frac{n^2(n+1)}{2} \).
Answer: Let the given statement be \( P(n) : 1 + 5 + 12 + 22 + 35 + \ldots + \frac{n(3n - 1)}{2} = \frac{n^2(n+1)}{2} \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( 1 \).
The RHS is \( \frac{1^2(1+1)}{2} = \frac{1 \cdot 2}{2} = 1 \).
Since LHS = RHS, \( P(1) \) is true. This proves the formula for the first term.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( 1 + 5 + 12 + \ldots + \frac{m(3m - 1)}{2} = \frac{m^2(m+1)}{2} \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( P(m+1) = \frac{(m+1)^2((m+1)+1)}{2} = \frac{(m+1)^2(m+2)}{2} \).
Consider the LHS for \( n = m+1 \):
\( \left( 1 + 5 + 12 + \ldots + \frac{m(3m - 1)}{2} \right) + \frac{(m+1)(3(m+1) - 1)}{2} \)
\( = \left( 1 + 5 + 12 + \ldots + \frac{m(3m - 1)}{2} \right) + \frac{(m+1)(3m + 3 - 1)}{2} \)
\( = \left( 1 + 5 + 12 + \ldots + \frac{m(3m - 1)}{2} \right) + \frac{(m+1)(3m + 2)}{2} \)
Using Equation (1) for the sum up to \( \frac{m(3m - 1)}{2} \):
\( = \frac{m^2(m+1)}{2} + \frac{(m+1)(3m + 2)}{2} \)
Take \( \frac{(m+1)}{2} \) as a common factor:
\( = \frac{(m+1)}{2} [m^2 + (3m + 2)] \)
\( = \frac{(m+1)}{2} [m^2 + 3m + 2] \)
Now, factor the quadratic term \( (m^2 + 3m + 2) \). This factors to \( (m+1)(m+2) \).
\( = \frac{(m+1)}{2} [(m+1)(m+2)] \)
\( = \frac{(m+1)^2(m+2)}{2} \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This guarantees the formula's validity for every natural number.
In simple words: We started by showing the formula works for the first term. Then, we assumed it works for any 'm' terms and proved it also works for 'm+1' terms. This step-by-step logic proves the formula for all natural numbers.

🎯 Exam Tip: When the terms of a series are given by a formula (like \( T_n = \frac{n(3n-1)}{2} \)), ensure you correctly substitute \( n=k+1 \) into this term before adding it to the sum for \( P(k) \).

Question 16. Prove by the method of mathematical induction that \( \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \frac{1}{10 \cdot 13} + \ldots + \frac{1}{(3n + 1)(3n + 4)} = \frac{n}{4(3n + 4)} \), for \( n \in N \).
Answer: Let the given statement be \( P(n) : \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \frac{1}{10 \cdot 13} + \ldots + \frac{1}{(3n + 1)(3n + 4)} = \frac{n}{4(3n + 4)} \).
Step 1: Base Case (n = 1)
For \( n = 1 \), the LHS is \( \frac{1}{4 \cdot 7} = \frac{1}{28} \).
The RHS is \( \frac{1}{4(3 \cdot 1 + 4)} = \frac{1}{4(7)} = \frac{1}{28} \).
Since LHS = RHS, \( P(1) \) is true. This verifies the formula for the initial term.
Step 2: Inductive Hypothesis
Assume that \( P(m) \) is true for some natural number \( m \).
So, \( \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots + \frac{1}{(3m + 1)(3m + 4)} = \frac{m}{4(3m + 4)} \). We will call this Equation (1).
Step 3: Inductive Step
Now, we need to prove that \( P(m+1) \) is true. This means we want to show that \( P(m+1) = \frac{(m+1)}{4(3(m+1) + 4)} = \frac{m+1}{4(3m + 7)} \).
Consider the LHS for \( n = m+1 \):
\( \left( \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots + \frac{1}{(3m + 1)(3m + 4)} \right) + \frac{1}{(3(m+1) + 1)(3(m+1) + 4)} \)
\( = \left( \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \ldots + \frac{1}{(3m + 1)(3m + 4)} \right) + \frac{1}{(3m + 4)(3m + 7)} \)
Using Equation (1) for the sum up to \( \frac{1}{(3m + 1)(3m + 4)} \):
\( = \frac{m}{4(3m + 4)} + \frac{1}{(3m + 4)(3m + 7)} \)
To combine, find a common denominator, which is \( 4(3m+4)(3m+7) \):
\( = \frac{m(3m + 7) + 4}{4(3m + 4)(3m + 7)} \)
\( = \frac{3m^2 + 7m + 4}{4(3m + 4)(3m + 7)} \)
Now, factor the quadratic term in the numerator \( (3m^2 + 7m + 4) \). We look for two numbers that multiply to \( 3 \times 4 = 12 \) and add to \( 7 \), which are 3 and 4.
\( 3m^2 + 7m + 4 = 3m^2 + 3m + 4m + 4 \)
\( = 3m(m + 1) + 4(m + 1) \)
\( = (3m + 4)(m + 1) \)
So, the expression becomes:
\( = \frac{(m + 1)(3m + 4)}{4(3m + 4)(3m + 7)} \)
Cancel out \( (3m + 4) \) from the numerator and denominator:
\( = \frac{m + 1}{4(3m + 7)} \). This is the RHS of \( P(n) \) with \( n \) replaced by \( m+1 \).
Therefore, \( P(m+1) \) is true.
Step 4: Conclusion
Since \( P(1) \) is true and \( P(m+1) \) is true whenever \( P(m) \) is true, by the Principle of Mathematical Induction, \( P(n) \) is true for all natural numbers \( n \in N \). This demonstrates the formula's consistency across all natural numbers.
In simple words: We first showed the formula works for the first term. Then, we assumed it works for any 'm' terms and used that to prove it also works for 'm+1' terms. This step-by-step logic confirms the formula is true for all natural numbers.

🎯 Exam Tip: Pay close attention to the denominator terms when combining fractions. Ensure you use the correct common denominator and factorize the numerator accurately to allow for proper simplification.

Question 17. Let \( S(k) = 1 + 3 + 5 + \ldots + (2k - 1) = 3 + k^2 \). Then which of the following is true?
(a) \( S(k) = S(k + 1) \)
(b) \( S(k) \neq S(k + 1) \)
(c) \( S(1) \) is correct
(d) Principle of mathematical induction can be used to prove the formula.
Answer: (b) \( S(k) \neq S(k + 1) \)
Consider the given statement \( S(k): 1 + 3 + 5 + \ldots + (2k - 1) = 3 + k^2 \).
Let's analyze each option:
(a) \( S(k) = S(k+1) \): The sum \( 1 + 3 + 5 + \ldots + (2k - 1) \) is clearly different from \( 1 + 3 + 5 + \ldots + (2k - 1) + (2(k+1) - 1) \). Also, \( 3+k^2 \) is different from \( 3+(k+1)^2 \). Therefore, this statement is false.
(b) \( S(k) \neq S(k+1) \): This is true. The sum of \( k \) odd numbers is \( k^2 \), and the sum of \( k+1 \) odd numbers is \( (k+1)^2 \). Since \( k^2 \neq (k+1)^2 \) for natural numbers, and the expression \( 3+k^2 \) itself changes for \( k+1 \), \( S(k) \) (meaning the expression for \( k \)) will not equal \( S(k+1) \) (meaning the expression for \( k+1 \)). Thus, \( S(k) \) will always be different from \( S(k+1) \) because adding another term changes the sum. This statement is true.
(c) \( S(1) \) is correct: For \( k=1 \), the Left Hand Side (LHS) of the statement \( S(k) \) is \( 1 \). The Right Hand Side (RHS) is \( 3 + 1^2 = 3 + 1 = 4 \). Since \( 1 \neq 4 \), the statement \( S(1) \) is false. So, this option is false.
(d) Principle of mathematical induction can be used to prove the formula: For mathematical induction to prove a formula, the base case must be true. As shown in (c), \( S(1) \) is false. Since the base case fails, the principle of mathematical induction cannot be used to prove this formula. Therefore, this option is false. Mathematical induction is only for proving true statements.
Based on this analysis, option (b) is the only true statement.
In simple words: The formula given is not actually true because it doesn't work for the first number (1). Since adding a new term always changes a sum, the sum for 'k' will always be different from the sum for 'k+1'. Because the basic check for 'n=1' fails, we cannot use mathematical induction to prove this formula. So, the statement that S(k) is not equal to S(k+1) is the only correct one.

🎯 Exam Tip: When evaluating statements about mathematical induction, always check the base case first. If the base case fails, the formula itself is incorrect, and induction cannot be applied to prove it, even if the inductive step works for the proposed (incorrect) formula.

Use the principle of mathematical induction to prove the following statements for all natural numbers \( n \).

Question 1. Use the principle of mathematical induction to prove that \( 1 + 2 + 3 + ... + n = \frac { 1 }{ 2 } n (n + 1) \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( 1 + 2 + 3 + ... + n = \frac { 1 }{ 2 } n (n + 1) \). This statement says that the sum of the first n natural numbers is given by the formula \( \frac { n (n + 1) }{ 2 } \).
Step 1: Base Case (n=1)
We check if the statement is true for the first natural number, \( n=1 \).
LHS (Left Hand Side) = \( 1 \)
RHS (Right Hand Side) = \( \frac { 1 }{ 2 } \times 1 \times (1 + 1) = \frac { 1 }{ 2 } \times 1 \times 2 = 1 \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(k) is true for some positive integer k. This means we assume:
\( 1 + 2 + 3 + ... + k = \frac { 1 }{ 2 } k (k + 1) \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(k+1) is also true. This means we need to show that:
\( 1 + 2 + 3 + ... + k + (k + 1) = \frac { 1 }{ 2 } (k + 1) ((k + 1) + 1) \)
Which simplifies to:
\( 1 + 2 + 3 + ... + k + (k + 1) = \frac { 1 }{ 2 } (k + 1) (k + 2) \)
Let's start with the LHS of P(k+1):
LHS = \( (1 + 2 + 3 + ... + k) + (k + 1) \)
Using our assumption from Equation (1), we can replace the sum of the first k terms:
LHS = \( \frac { 1 }{ 2 } k (k + 1) + (k + 1) \)
Now, we can factor out the common term \( (k + 1) \):
LHS = \( (k + 1) \left( \frac { k }{ 2 } + 1 \right) \)
To combine the terms inside the parenthesis, find a common denominator:
LHS = \( (k + 1) \left( \frac { k + 2 }{ 2 } \right) \)
LHS = \( \frac { 1 }{ 2 } (k + 1) (k + 2) \)
This is exactly the RHS of P(k+1). So, P(k+1) is true.
Conclusion:
Since P(1) is true and P(k) implies P(k+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We first checked if the formula works for the very first number (1). Then, we assumed it works for any number 'k' and used that to show it must also work for the next number 'k+1'. Because it starts true and stays true, it works for all natural numbers.

🎯 Exam Tip: Always clearly state your P(n) statement, perform the base case check, articulate the inductive hypothesis, and show the inductive step by manipulating the LHS of P(k+1) to match its RHS using P(k). This structured approach earns full marks.

Question 2. Prove by mathematical induction that \( 2 + 4 + 6 + ... + 2n = n (n + 1) \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( 2 + 4 + 6 + ... + 2n = n (n + 1) \). This statement shows that the sum of the first n even numbers is equal to \( n(n+1) \).
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( 2 \)
RHS = \( 1 (1 + 1) = 1 \times 2 = 2 \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( 2 + 4 + 6 + ... + 2m = m (m + 1) \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( 2 + 4 + 6 + ... + 2m + 2(m + 1) = (m + 1) ((m + 1) + 1) \)
Which simplifies to:
\( 2 + 4 + 6 + ... + 2m + 2(m + 1) = (m + 1) (m + 2) \)
Let's start with the LHS of P(m+1):
LHS = \( (2 + 4 + 6 + ... + 2m) + 2(m + 1) \)
Using our assumption from Equation (1):
LHS = \( m (m + 1) + 2 (m + 1) \)
Factor out the common term \( (m + 1) \):
LHS = \( (m + 1) (m + 2) \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We showed that the formula works for the first number, 1. Then, we showed that if it works for any number 'm', it will also work for the next number 'm+1'. This means the formula is true for all whole numbers.

🎯 Exam Tip: When performing the inductive step, clearly identify the sum of the first 'm' terms so you can substitute the inductive hypothesis (Equation 1) directly. This keeps your steps organized and easy to follow.

Question 3. Prove by mathematical induction that \( 1^2 + 2^2 + 3^2 + ... + n^2 = \frac { 1 }{ 6 }n (n + 1) (2n + 1) \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( 1^2 + 2^2 + 3^2 + ... + n^2 = \frac { 1 }{ 6 } n (n + 1) (2n + 1) \). This formula gives the sum of the squares of the first n natural numbers.
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( 1^2 = 1 \)
RHS = \( \frac { 1 }{ 6 } \times 1 \times (1 + 1) \times (2 \times 1 + 1) = \frac { 1 }{ 6 } \times 1 \times 2 \times 3 = \frac { 6 }{ 6 } = 1 \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( 1^2 + 2^2 + 3^2 + ... + m^2 = \frac { 1 }{ 6 } m (m + 1) (2m + 1) \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( 1^2 + 2^2 + ... + m^2 + (m + 1)^2 = \frac { 1 }{ 6 } (m + 1) ((m + 1) + 1) (2(m + 1) + 1) \)
Which simplifies to:
\( 1^2 + 2^2 + ... + m^2 + (m + 1)^2 = \frac { 1 }{ 6 } (m + 1) (m + 2) (2m + 3) \)
Let's start with the LHS of P(m+1):
LHS = \( (1^2 + 2^2 + ... + m^2) + (m + 1)^2 \)
Using our assumption from Equation (1):
LHS = \( \frac { 1 }{ 6 } m (m + 1) (2m + 1) + (m + 1)^2 \)
Factor out the common term \( (m + 1) \):
LHS = \( (m + 1) \left[ \frac { m (2m + 1) }{ 6 } + (m + 1) \right] \)
To combine the terms inside the square brackets, find a common denominator:
LHS = \( (m + 1) \left[ \frac { 2m^2 + m + 6(m + 1) }{ 6 } \right] \)
LHS = \( (m + 1) \left[ \frac { 2m^2 + m + 6m + 6 }{ 6 } \right] \)
LHS = \( (m + 1) \left[ \frac { 2m^2 + 7m + 6 }{ 6 } \right] \)
Now, we factor the quadratic expression \( 2m^2 + 7m + 6 \). We look for two numbers that multiply to \( 2 \times 6 = 12 \) and add up to 7, which are 3 and 4.
\( 2m^2 + 7m + 6 = 2m^2 + 4m + 3m + 6 = 2m(m + 2) + 3(m + 2) = (2m + 3)(m + 2) \)
So, substitute this back into the LHS:
LHS = \( \frac { 1 }{ 6 } (m + 1) (m + 2) (2m + 3) \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We check the formula for the first number (1). Then we assume it holds for 'm' and use this to prove it also holds for 'm+1'. This confirms the formula works for all natural numbers because the sum of squares follows a predictable pattern.

🎯 Exam Tip: Factoring quadratic expressions correctly in the inductive step is crucial. Double-check your factorization by expanding it back to ensure it matches the original quadratic. This is a common place for errors.

Question 4. Prove by mathematical induction that \( 1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = \frac { 1 }{ 3 } n (2n - 1) (2n + 1) \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( 1^2 + 3^2 + 5^2 + ... + (2n - 1)^2 = \frac { 1 }{ 3 } n (2n - 1) (2n + 1) \). This formula calculates the sum of the squares of the first n odd numbers.
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( (2 \times 1 - 1)^2 = 1^2 = 1 \)
RHS = \( \frac { 1 }{ 3 } \times 1 \times (2 \times 1 - 1) \times (2 \times 1 + 1) = \frac { 1 }{ 3 } \times 1 \times 1 \times 3 = 1 \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( 1^2 + 3^2 + 5^2 + ... + (2m - 1)^2 = \frac { 1 }{ 3 } m (2m - 1) (2m + 1) \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( 1^2 + 3^2 + ... + (2m - 1)^2 + (2(m + 1) - 1)^2 = \frac { 1 }{ 3 } (m + 1) (2(m + 1) - 1) (2(m + 1) + 1) \)
Which simplifies to:
\( 1^2 + 3^2 + ... + (2m - 1)^2 + (2m + 1)^2 = \frac { 1 }{ 3 } (m + 1) (2m + 1) (2m + 3) \)
Let's start with the LHS of P(m+1):
LHS = \( (1^2 + 3^2 + ... + (2m - 1)^2) + (2m + 1)^2 \)
Using our assumption from Equation (1):
LHS = \( \frac { 1 }{ 3 } m (2m - 1) (2m + 1) + (2m + 1)^2 \)
Factor out the common term \( (2m + 1) \):
LHS = \( (2m + 1) \left[ \frac { m (2m - 1) }{ 3 } + (2m + 1) \right] \)
To combine the terms inside the square brackets, find a common denominator:
LHS = \( (2m + 1) \left[ \frac { 2m^2 - m + 3(2m + 1) }{ 3 } \right] \)
LHS = \( (2m + 1) \left[ \frac { 2m^2 - m + 6m + 3 }{ 3 } \right] \)
LHS = \( (2m + 1) \left[ \frac { 2m^2 + 5m + 3 }{ 3 } \right] \)
Now, we factor the quadratic expression \( 2m^2 + 5m + 3 \). We look for two numbers that multiply to \( 2 \times 3 = 6 \) and add up to 5, which are 2 and 3.
\( 2m^2 + 5m + 3 = 2m^2 + 2m + 3m + 3 = 2m(m + 1) + 3(m + 1) = (2m + 3)(m + 1) \)
So, substitute this back into the LHS:
LHS = \( \frac { 1 }{ 3 } (2m + 1) (m + 1) (2m + 3) \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We check the formula for the first odd number, which is 1. Then we assume it holds for 'm' odd numbers and use this to prove it also holds for 'm+1' odd numbers. This confirms the formula works for all sums of squares of odd numbers.

🎯 Exam Tip: Pay close attention to the term representing \( (k+1)^{th} \) element in your series; for odd numbers, it's \( (2(k+1)-1)^2 = (2k+1)^2 \). Ensure you add the correct next term to the LHS in the inductive step.

Question 5. Prove by mathematical induction that \( 2 + 2^2 + 2^3 + ... + 2^n = 2 (2^n - 1) \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( 2 + 2^2 + 2^3 + ... + 2^n = 2 (2^n - 1) \). This formula gives the sum of the first n terms of a geometric series where the first term is 2 and the common ratio is 2.
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( 2^1 = 2 \)
RHS = \( 2 (2^1 - 1) = 2 (2 - 1) = 2 \times 1 = 2 \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( 2 + 2^2 + 2^3 + ... + 2^m = 2 (2^m - 1) \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( 2 + 2^2 + ... + 2^m + 2^{m+1} = 2 (2^{(m+1)} - 1) \)
Let's start with the LHS of P(m+1):
LHS = \( (2 + 2^2 + ... + 2^m) + 2^{m+1} \)
Using our assumption from Equation (1):
LHS = \( 2 (2^m - 1) + 2^{m+1} \)
Expand the first term:
LHS = \( 2 \cdot 2^m - 2 + 2^{m+1} \)
Since \( 2 \cdot 2^m = 2^{m+1} \):
LHS = \( 2^{m+1} - 2 + 2^{m+1} \)
Combine the \( 2^{m+1} \) terms:
LHS = \( 2 \cdot 2^{m+1} - 2 \)
Factor out 2:
LHS = \( 2 (2^{m+1} - 1) \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We verify the formula for the first term. Then we show that if the formula works for 'm' terms, it will also work correctly for 'm+1' terms. This confirms that the sum of powers of 2 follows this formula for all natural numbers.

🎯 Exam Tip: When dealing with powers, remember the rules of exponents, especially \( a^x \cdot a^y = a^{x+y} \). This helps in simplifying expressions like \( 2 \cdot 2^m \) to \( 2^{m+1} \) and combining terms correctly.

Question 6. Prove by mathematical induction that \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + n (n + 1) = \frac{n(n+1)(n+2)}{3} \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + n (n + 1) = \frac{n(n+1)(n+2)}{3} \). This formula sums the products of consecutive natural numbers.
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( 1 \cdot 2 = 2 \)
RHS = \( \frac{1(1+1)(1+2)}{3} = \frac{1 \times 2 \times 3}{3} = \frac{6}{3} = 2 \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( 1 \cdot 2 + 2 \cdot 3 + ... + m (m + 1) = \frac{m(m+1)(m+2)}{3} \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( 1 \cdot 2 + 2 \cdot 3 + ... + m (m + 1) + (m + 1)((m + 1) + 1) = \frac{(m+1)((m+1)+1)((m+1)+2)}{3} \)
Which simplifies to:
\( 1 \cdot 2 + 2 \cdot 3 + ... + m (m + 1) + (m + 1)(m + 2) = \frac{(m+1)(m+2)(m+3)}{3} \)
Let's start with the LHS of P(m+1):
LHS = \( (1 \cdot 2 + 2 \cdot 3 + ... + m (m + 1)) + (m + 1)(m + 2) \)
Using our assumption from Equation (1):
LHS = \( \frac{m(m+1)(m+2)}{3} + (m + 1)(m + 2) \)
Factor out the common term \( (m + 1)(m + 2) \):
LHS = \( (m + 1)(m + 2) \left[ \frac{m}{3} + 1 \right] \)
To combine the terms inside the square brackets, find a common denominator:
LHS = \( (m + 1)(m + 2) \left[ \frac{m + 3}{3} \right] \)
LHS = \( \frac{(m+1)(m+2)(m+3)}{3} \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We checked if the sum of consecutive number products works for the first pair (1 times 2). Then we showed that if it works for 'm' pairs, it will also work for 'm+1' pairs. This means the formula holds true for all natural numbers.

🎯 Exam Tip: Look for common factors to simplify the expression after substituting the inductive hypothesis. This often involves taking out `(m+1)` and `(m+2)` terms as done here, making the remaining algebra much simpler.

Question 7. Prove by mathematical induction that \( 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + ... + n \cdot 2^n = (n - 1) 2^{n+1} + 2 \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + ... + n \cdot 2^n = (n - 1) 2^{n+1} + 2 \). This formula relates to the sum of a series involving products of natural numbers and powers of 2.
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( 1 \cdot 2^1 = 2 \)
RHS = \( (1 - 1) 2^{1+1} + 2 = 0 \cdot 2^2 + 2 = 0 + 2 = 2 \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( 1 \cdot 2 + 2 \cdot 2^2 + 3 \cdot 2^3 + ... + m \cdot 2^m = (m - 1) 2^{m+1} + 2 \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( 1 \cdot 2 + 2 \cdot 2^2 + ... + m \cdot 2^m + (m+1) 2^{m+1} = ((m+1) - 1) 2^{((m+1)+1)} + 2 \)
Which simplifies to:
\( 1 \cdot 2 + 2 \cdot 2^2 + ... + m \cdot 2^m + (m+1) 2^{m+1} = m \cdot 2^{m+2} + 2 \)
Let's start with the LHS of P(m+1):
LHS = \( (1 \cdot 2 + 2 \cdot 2^2 + ... + m \cdot 2^m) + (m+1) 2^{m+1} \)
Using our assumption from Equation (1):
LHS = \( (m - 1) 2^{m+1} + 2 + (m+1) 2^{m+1} \)
Group the terms with \( 2^{m+1} \):
LHS = \( ((m - 1) + (m + 1)) 2^{m+1} + 2 \)
Simplify the expression inside the parenthesis:
LHS = \( (2m) 2^{m+1} + 2 \)
Rewrite \( 2m \cdot 2^{m+1} \) using exponent rules:
LHS = \( m \cdot (2 \cdot 2^{m+1}) + 2 \)
LHS = \( m \cdot 2^{m+2} + 2 \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We showed that the formula holds for the first number (n=1). Then we assumed it works for any number 'm' and used that to prove it also works for 'm+1'. This means the formula is always true for all natural numbers.

🎯 Exam Tip: Be careful with algebraic manipulation involving exponents. Remember to combine terms with the same base and exponent correctly. A common mistake is misapplying the rules of exponents, especially when adding or factoring terms.

Question 8. Prove by mathematical induction that \( 3 \cdot 2^2 + 3^2 \cdot 2^3 + 3^3 \cdot 2^4 + ... + 3^n \cdot 2^{n+1} = \frac { 12 }{ 5 }(6^n - 1) \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( 3 \cdot 2^2 + 3^2 \cdot 2^3 + 3^3 \cdot 2^4 + ... + 3^n \cdot 2^{n+1} = \frac { 12 }{ 5 }(6^n - 1) \). This formula sums a series where each term is a product of powers of 3 and powers of 2.
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( 3^1 \cdot 2^{1+1} = 3 \cdot 2^2 = 3 \cdot 4 = 12 \)
RHS = \( \frac { 12 }{ 5 }(6^1 - 1) = \frac { 12 }{ 5 }(6 - 1) = \frac { 12 }{ 5 } \times 5 = 12 \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( 3 \cdot 2^2 + 3^2 \cdot 2^3 + ... + 3^m \cdot 2^{m+1} = \frac { 12 }{ 5 }(6^m - 1) \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( 3 \cdot 2^2 + ... + 3^m \cdot 2^{m+1} + 3^{m+1} \cdot 2^{(m+1)+1} = \frac { 12 }{ 5 }(6^{m+1} - 1) \)
Which simplifies to:
\( 3 \cdot 2^2 + ... + 3^m \cdot 2^{m+1} + 3^{m+1} \cdot 2^{m+2} = \frac { 12 }{ 5 }(6^{m+1} - 1) \)
Let's start with the LHS of P(m+1):
LHS = \( (3 \cdot 2^2 + ... + 3^m \cdot 2^{m+1}) + 3^{m+1} \cdot 2^{m+2} \)
Using our assumption from Equation (1):
LHS = \( \frac { 12 }{ 5 }(6^m - 1) + 3^{m+1} \cdot 2^{m+2} \)
Rewrite the second term using exponent rules to get base 6:
\( 3^{m+1} \cdot 2^{m+2} = (3^m \cdot 3^1) \cdot (2^m \cdot 2^2) = (3^m \cdot 2^m) \cdot (3 \cdot 2^2) = 6^m \cdot 3 \cdot 4 = 12 \cdot 6^m \)
Substitute this back into the LHS:
LHS = \( \frac { 12 }{ 5 }(6^m - 1) + 12 \cdot 6^m \)
LHS = \( \frac { 12 \cdot 6^m }{ 5 } - \frac { 12 }{ 5 } + 12 \cdot 6^m \)
Group the terms with \( 6^m \):
LHS = \( 12 \cdot 6^m \left( \frac{1}{5} + 1 \right) - \frac{12}{5} \)
LHS = \( 12 \cdot 6^m \left( \frac{1+5}{5} \right) - \frac{12}{5} \)
LHS = \( 12 \cdot 6^m \left( \frac{6}{5} \right) - \frac{12}{5} \)
LHS = \( \frac{12}{5} \cdot 6^{m+1} - \frac{12}{5} \)
Factor out \( \frac{12}{5} \):
LHS = \( \frac{12}{5} (6^{m+1} - 1) \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We checked the formula for the first term. Then we assumed it works for 'm' terms and used this to show it also works for 'm+1' terms. This confirms the formula correctly sums the series for all natural numbers.

🎯 Exam Tip: When terms involve products of different powers (like \( 3^n \cdot 2^{n+1} \)), try to express them with a common base if possible (like \( 6^n \)). This helps simplify calculations greatly and matches the structure of the RHS.

Question 9. Prove by mathematical induction that \( 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + ... + (2n - 1) (2n + 1) = \frac{n(4 n^2+6 n-1)}{3} \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( 1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + ... + (2n - 1) (2n + 1) = \frac{n(4 n^2+6 n-1)}{3} \). This formula calculates the sum of the products of consecutive odd numbers.
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( (2 \times 1 - 1) (2 \times 1 + 1) = 1 \cdot 3 = 3 \)
RHS = \( \frac{1(4 \cdot 1^2+6 \cdot 1-1)}{3} = \frac{1(4+6-1)}{3} = \frac{9}{3} = 3 \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( 1 \cdot 3 + 3 \cdot 5 + ... + (2m - 1) (2m + 1) = \frac{m(4 m^2+6 m-1)}{3} \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( 1 \cdot 3 + ... + (2m - 1) (2m + 1) + (2(m+1) - 1) (2(m+1) + 1) = \frac{(m+1)(4 (m+1)^2+6(m+1)-1)}{3} \)
Which simplifies to:
\( 1 \cdot 3 + ... + (2m - 1) (2m + 1) + (2m + 1) (2m + 3) = \frac{(m+1)(4 m^2+14 m+9)}{3} \)
Let's start with the LHS of P(m+1):
LHS = \( (1 \cdot 3 + ... + (2m - 1) (2m + 1)) + (2m + 1) (2m + 3) \)
Using our assumption from Equation (1):
LHS = \( \frac{m(4 m^2+6 m-1)}{3} + (2m + 1) (2m + 3) \)
To combine, find a common denominator:
LHS = \( \frac{m(4 m^2+6 m-1) + 3(2m + 1)(2m + 3)}{3} \)
Expand the terms in the numerator:
\( m(4 m^2+6 m-1) = 4m^3 + 6m^2 - m \)
\( 3(2m + 1)(2m + 3) = 3(4m^2 + 6m + 2m + 3) = 3(4m^2 + 8m + 3) = 12m^2 + 24m + 9 \)
Substitute these expanded terms back into the LHS:
LHS = \( \frac{4m^3 + 6m^2 - m + 12m^2 + 24m + 9}{3} \)
LHS = \( \frac{4m^3 + 18m^2 + 23m + 9}{3} \)
Now, let's verify if this matches the RHS for P(m+1). The RHS is \( \frac{(m+1)(4(m+1)^2+6(m+1)-1)}{3} \).
First, expand the quadratic part: \( 4(m+1)^2+6(m+1)-1 = 4(m^2+2m+1)+6m+6-1 = 4m^2+8m+4+6m+5 = 4m^2+14m+9 \).
So, the RHS is \( \frac{(m+1)(4m^2+14m+9)}{3} \).
Now, expand the numerator of the RHS:
\( (m+1)(4m^2+14m+9) = m(4m^2+14m+9) + 1(4m^2+14m+9) \)
\( = 4m^3+14m^2+9m + 4m^2+14m+9 \)
\( = 4m^3+18m^2+23m+9 \)
Since the derived LHS matches the expanded RHS, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We proved that this formula, which adds up products of odd numbers, works for the first pair. Then we showed that if it works for 'm' pairs, it automatically works for 'm+1' pairs. So, it's true for all natural numbers.

🎯 Exam Tip: Questions involving complex algebraic expressions in the inductive step often require careful expansion and factorization. It's helpful to simplify the target RHS separately to ensure your derived LHS matches perfectly.

Question 10. Prove by the method of mathematical induction that:
(i) \( a + (a + d) + (a + 2d) + ... + [a + (n - 1) d] = \frac { n }{ 2 } [2a + (n - 1) d] \)
(ii) \( a + ar + ar^2 + ... + ar^{n-1} = \frac{a(1-r^n)}{1-r} \), where \( r \neq 1 \).
Answer:
(i) Let P(n) be the statement: \( a + (a + d) + (a + 2d) + ... + [a + (n - 1) d] = \frac { n }{ 2 } [2a + (n - 1) d] \). This formula calculates the sum of an arithmetic progression (AP).
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( a + (1 - 1)d = a \)
RHS = \( \frac { 1 }{ 2 } [2a + (1 - 1) d] = \frac { 1 }{ 2 } [2a + 0] = a \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( a + (a + d) + ... + [a + (m - 1) d] = \frac { m }{ 2 } [2a + (m - 1) d] \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( a + (a + d) + ... + [a + (m - 1) d] + [a + ((m+1) - 1) d] = \frac { m+1 }{ 2 } [2a + ((m+1) - 1) d] \)
Which simplifies to:
\( a + (a + d) + ... + [a + (m - 1) d] + (a + md) = \frac { m+1 }{ 2 } [2a + md] \)
Let's start with the LHS of P(m+1):
LHS = \( (a + (a + d) + ... + [a + (m - 1) d]) + (a + md) \)
Using our assumption from Equation (1):
LHS = \( \frac { m }{ 2 } [2a + (m - 1) d] + (a + md) \)
Multiply the terms within the bracket and find a common denominator:
LHS = \( \frac { m(2a + md - d) + 2(a + md) }{ 2 } \)
LHS = \( \frac { 2am + m^2d - md + 2a + 2md }{ 2 } \)
Combine like terms:
LHS = \( \frac { 2am + 2a + m^2d + md }{ 2 } \)
Factor out common terms:
LHS = \( \frac { 2a(m + 1) + md(m + 1) }{ 2 } \)
LHS = \( \frac { (m + 1)(2a + md) }{ 2 } \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.

(ii) Let P(n) be the statement: \( a + ar + ar^2 + ... + ar^{n-1} = \frac{a(1-r^n)}{1-r} \) for \( r \neq 1 \). This formula calculates the sum of a geometric progression (GP).
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( ar^{1-1} = ar^0 = a \)
RHS = \( \frac{a(1-r^1)}{1-r} = \frac{a(1-r)}{1-r} = a \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( a + ar + ar^2 + ... + ar^{m-1} = \frac{a(1-r^m)}{1-r} \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( a + ar + ar^2 + ... + ar^{m-1} + ar^{(m+1)-1} = \frac{a(1-r^{m+1})}{1-r} \)
Which simplifies to:
\( a + ar + ar^2 + ... + ar^{m-1} + ar^m = \frac{a(1-r^{m+1})}{1-r} \)
Let's start with the LHS of P(m+1):
LHS = \( (a + ar + ar^2 + ... + ar^{m-1}) + ar^m \)
Using our assumption from Equation (1):
LHS = \( \frac{a(1-r^m)}{1-r} + ar^m \)
To combine, find a common denominator:
LHS = \( \frac{a(1-r^m) + ar^m(1-r)}{1-r} \)
Expand the terms in the numerator:
LHS = \( \frac{a - ar^m + ar^m - ar^m \cdot r}{1-r} \)
LHS = \( \frac{a - ar^{m+1}}{1-r} \)
Factor out \( a \):
LHS = \( \frac{a(1 - r^{m+1})}{1-r} \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion for both parts:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, both statements P(n) are true for all natural numbers \( n \) (with \( r \neq 1 \) for the second part).
In simple words: For both parts, we showed that the formula works for the first term. Then we assumed it works for 'm' terms and used that to prove it also works for 'm+1' terms. This means both formulas correctly calculate the sums for arithmetic and geometric progressions for all natural numbers.

🎯 Exam Tip: When proving sum formulas for AP or GP, ensure you correctly identify the common difference 'd' or common ratio 'r' and the nth term. The key is to correctly express the \( (m+1)^{th} \) term and combine it with the sum of the first 'm' terms using the inductive hypothesis.

Question 11. Prove by mathematical induction that \( 5 + 15 + 45 + ... + 5 \cdot 3^{n-1} = \frac { 5 }{ 2 } (3^n - 1) \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( 5 + 15 + 45 + ... + 5 \cdot 3^{n-1} = \frac { 5 }{ 2 } (3^n - 1) \). This is a geometric series where the first term is 5 and the common ratio is 3.
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( 5 \cdot 3^{1-1} = 5 \cdot 3^0 = 5 \cdot 1 = 5 \)
RHS = \( \frac { 5 }{ 2 } (3^1 - 1) = \frac { 5 }{ 2 } (3 - 1) = \frac { 5 }{ 2 } \times 2 = 5 \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( 5 + 15 + 45 + ... + 5 \cdot 3^{m-1} = \frac { 5 }{ 2 } (3^m - 1) \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( 5 + 15 + ... + 5 \cdot 3^{m-1} + 5 \cdot 3^{(m+1)-1} = \frac { 5 }{ 2 } (3^{m+1} - 1) \)
Which simplifies to:
\( 5 + 15 + ... + 5 \cdot 3^{m-1} + 5 \cdot 3^m = \frac { 5 }{ 2 } (3^{m+1} - 1) \)
Let's start with the LHS of P(m+1):
LHS = \( (5 + 15 + ... + 5 \cdot 3^{m-1}) + 5 \cdot 3^m \)
Using our assumption from Equation (1):
LHS = \( \frac { 5 }{ 2 } (3^m - 1) + 5 \cdot 3^m \)
To combine, find a common denominator:
LHS = \( \frac { 5(3^m - 1) + 2 \cdot 5 \cdot 3^m }{ 2 } \)
LHS = \( \frac { 5 \cdot 3^m - 5 + 10 \cdot 3^m }{ 2 } \)
Combine the terms with \( 3^m \):
LHS = \( \frac { (5 \cdot 3^m + 10 \cdot 3^m) - 5 }{ 2 } \)
LHS = \( \frac { 15 \cdot 3^m - 5 }{ 2 } \)
Rewrite 15 as \( 5 \cdot 3 \):
LHS = \( \frac { 5 \cdot 3 \cdot 3^m - 5 }{ 2 } \)
LHS = \( \frac { 5 \cdot 3^{m+1} - 5 }{ 2 } \)
Factor out 5:
LHS = \( \frac { 5 (3^{m+1} - 1) }{ 2 } \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We checked the formula for the first term of this multiplying series. Then, we assumed it works for 'm' terms and used this to show it also works for 'm+1' terms. This proves that the formula is always correct for this series.

🎯 Exam Tip: When dealing with geometric series, make sure to correctly identify the first term and the common ratio. This helps you write down the \( (m+1)^{th} \) term accurately and simplify expressions using exponent rules.

Question 12. Prove by mathematical induction that \( \frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\frac{1}{7 \cdot 9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)} \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( \frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\frac{1}{7 \cdot 9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)} \). This formula sums a series of fractions where each denominator is a product of consecutive odd numbers.
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( \frac{1}{(2 \cdot 1+1)(2 \cdot 1+3)} = \frac{1}{3 \cdot 5} = \frac{1}{15} \)
RHS = \( \frac{1}{3(2 \cdot 1+3)} = \frac{1}{3(5)} = \frac{1}{15} \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( \frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\ldots+\frac{1}{(2m+1)(2m+3)}=\frac{m}{3(2m+3)} \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( \frac{1}{3 \cdot 5}+\ldots+\frac{1}{(2m+1)(2m+3)} + \frac{1}{(2(m+1)+1)(2(m+1)+3)} = \frac{m+1}{3(2(m+1)+3)} \)
Which simplifies to:
\( \frac{1}{3 \cdot 5}+\ldots+\frac{1}{(2m+1)(2m+3)} + \frac{1}{(2m+3)(2m+5)} = \frac{m+1}{3(2m+5)} \)
Let's start with the LHS of P(m+1):
LHS = \( \left( \frac{1}{3 \cdot 5}+\ldots+\frac{1}{(2m+1)(2m+3)} \right) + \frac{1}{(2m+3)(2m+5)} \)
Using our assumption from Equation (1):
LHS = \( \frac{m}{3(2m+3)} + \frac{1}{(2m+3)(2m+5)} \)
To combine these fractions, find a common denominator, which is \( 3(2m+3)(2m+5) \):
LHS = \( \frac{m(2m+5) + 3}{3(2m+3)(2m+5)} \)
Expand the numerator:
LHS = \( \frac{2m^2+5m+3}{3(2m+3)(2m+5)} \)
Now, factor the quadratic expression \( 2m^2+5m+3 \). We found in Question 4 that this factors as \( (2m+3)(m+1) \).
So, substitute this back into the LHS:
LHS = \( \frac{(2m+3)(m+1)}{3(2m+3)(2m+5)} \)
Cancel out the common term \( (2m+3) \), assuming \( 2m+3 \neq 0 \) (which is true for natural numbers m):
LHS = \( \frac{m+1}{3(2m+5)} \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We tested the formula for the first term, which matched. Then, we assumed the formula works for 'm' terms and used that to prove it also works for 'm+1' terms. This shows the formula is correct for adding up all terms in this sequence of fractions.

🎯 Exam Tip: For sums of fractions involving products in the denominator, partial fraction decomposition is often helpful conceptually, but for induction, directly finding a common denominator and factoring the numerator is usually the most straightforward path.

Question 13. Prove by mathematical induction that \( \frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1} \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( \frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1} \). This formula sums a series of fractions where each denominator is a product of terms from an arithmetic progression.
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( \frac{1}{(3 \cdot 1 - 2)(3 \cdot 1 + 1)} = \frac{1}{1 \cdot 4} = \frac{1}{4} \)
RHS = \( \frac{1}{3 \cdot 1 + 1} = \frac{1}{4} \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( \frac{1}{1 \cdot 4}+\frac{1}{4 \cdot 7}+\ldots+\frac{1}{(3m-2)(3m+1)}=\frac{m}{3m+1} \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( \frac{1}{1 \cdot 4}+\ldots+\frac{1}{(3m-2)(3m+1)} + \frac{1}{(3(m+1)-2)(3(m+1)+1)} = \frac{m+1}{3(m+1)+1} \)
Which simplifies to:
\( \frac{1}{1 \cdot 4}+\ldots+\frac{1}{(3m-2)(3m+1)} + \frac{1}{(3m+1)(3m+4)} = \frac{m+1}{3m+4} \)
Let's start with the LHS of P(m+1):
LHS = \( \left( \frac{1}{1 \cdot 4}+\ldots+\frac{1}{(3m-2)(3m+1)} \right) + \frac{1}{(3m+1)(3m+4)} \)
Using our assumption from Equation (1):
LHS = \( \frac{m}{3m+1} + \frac{1}{(3m+1)(3m+4)} \)
To combine these fractions, find a common denominator, which is \( (3m+1)(3m+4) \):
LHS = \( \frac{m(3m+4) + 1}{(3m+1)(3m+4)} \)
Expand the numerator:
LHS = \( \frac{3m^2+4m+1}{(3m+1)(3m+4)} \)
Now, factor the quadratic expression \( 3m^2+4m+1 \). We look for two numbers that multiply to \( 3 \times 1 = 3 \) and add up to 4, which are 1 and 3.
\( 3m^2+4m+1 = 3m^2+3m+m+1 = 3m(m+1) + 1(m+1) = (3m+1)(m+1) \)
So, substitute this back into the LHS:
LHS = \( \frac{(3m+1)(m+1)}{(3m+1)(3m+4)} \)
Cancel out the common term \( (3m+1) \), assuming \( 3m+1 \neq 0 \) (which is true for natural numbers m):
LHS = \( \frac{m+1}{3m+4} \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We checked the formula for the first term of the series, and it matched. Then, we showed that if we assume it works for 'm' terms, it will also work for 'm+1' terms. This means the formula is correct for adding up all terms in this sequence of fractions.

🎯 Exam Tip: Be sure to correctly find the general \( n^{th} \) term of the series to formulate P(n) accurately. In series with products in the denominator, the terms often form a specific pattern like an AP, which helps identify the next term. Remember to simplify the numerator by factoring after finding a common denominator.

Question 14. Prove by mathematical induction that \( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^n}=1-\frac{1}{2^n} \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^n}=1-\frac{1}{2^n} \). This formula calculates the sum of a geometric series where the first term is \( \frac{1}{2} \) and the common ratio is \( \frac{1}{2} \).
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( \frac{1}{2^1} = \frac{1}{2} \)
RHS = \( 1 - \frac{1}{2^1} = 1 - \frac{1}{2} = \frac{1}{2} \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( \frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^m}=1-\frac{1}{2^m} \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( \frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^m} + \frac{1}{2^{m+1}} = 1-\frac{1}{2^{m+1}} \)
Let's start with the LHS of P(m+1):
LHS = \( \left( \frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^m} \right) + \frac{1}{2^{m+1}} \)
Using our assumption from Equation (1):
LHS = \( \left(1-\frac{1}{2^m}\right) + \frac{1}{2^{m+1}} \)
Rewrite \( \frac{1}{2^{m+1}} \) as \( \frac{1}{2 \cdot 2^m} \):
LHS = \( 1-\frac{1}{2^m} + \frac{1}{2 \cdot 2^m} \)
Factor out \( -\frac{1}{2^m} \):
LHS = \( 1 - \frac{1}{2^m} \left(1 - \frac{1}{2}\right) \)
Simplify the expression in the parenthesis:
LHS = \( 1 - \frac{1}{2^m} \left(\frac{1}{2}\right) \)
LHS = \( 1 - \frac{1}{2^{m+1}} \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We checked the formula for the first term (1/2), and it worked. Then, we assumed it works for 'm' terms and showed it must also work for 'm+1' terms. This proves that the formula for summing powers of 1/2 is always correct.

🎯 Exam Tip: When dealing with sums of fractions with powers in the denominator, careful manipulation of exponents and finding common factors/denominators is key. Ensure you correctly factor out \( \frac{1}{2^m} \) to simplify the expression efficiently.

Question 15. Prove by mathematical induction that \( 1 + 5 + 12 + 22 + 35 + ... + \frac { n }{ 2 } (3n - 1) = \frac{n^2(n+1)}{2} \) for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( 1 + 5 + 12 + 22 + 35 + ... + \frac { n }{ 2 } (3n - 1) = \frac{n^2(n+1)}{2} \). This formula sums a series where the \( n^{th} \) term is given by \( \frac{n(3n-1)}{2} \).
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( \frac { 1 }{ 2 } (3 \cdot 1 - 1) = \frac { 1 }{ 2 } (2) = 1 \)
RHS = \( \frac{1^2(1+1)}{2} = \frac{1 \cdot 2}{2} = 1 \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( 1 + 5 + 12 + ... + \frac { m }{ 2 } (3m - 1) = \frac{m^2(m+1)}{2} \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( 1 + 5 + ... + \frac { m }{ 2 } (3m - 1) + \frac { m+1 }{ 2 } (3(m+1) - 1) = \frac{(m+1)^2((m+1)+1)}{2} \)
Which simplifies to:
\( 1 + 5 + ... + \frac { m }{ 2 } (3m - 1) + \frac { m+1 }{ 2 } (3m + 2) = \frac{(m+1)^2(m+2)}{2} \)
Let's start with the LHS of P(m+1):
LHS = \( \left( 1 + 5 + ... + \frac { m }{ 2 } (3m - 1) \right) + \frac { m+1 }{ 2 } (3m + 2) \)
Using our assumption from Equation (1):
LHS = \( \frac{m^2(m+1)}{2} + \frac{(m+1)(3m+2)}{2} \)
Factor out the common term \( \frac{m+1}{2} \):
LHS = \( \frac{m+1}{2} [m^2 + (3m + 2)] \)
Simplify the expression inside the square brackets:
LHS = \( \frac{m+1}{2} [m^2 + 3m + 2] \)
Factor the quadratic expression \( m^2 + 3m + 2 \). We look for two numbers that multiply to 2 and add up to 3, which are 1 and 2.
\( m^2 + 3m + 2 = (m+1)(m+2) \)
So, substitute this back into the LHS:
LHS = \( \frac{m+1}{2} (m+1)(m+2) \)
LHS = \( \frac{(m+1)^2(m+2)}{2} \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We checked the formula for the first term, and it worked. Then, we assumed it works for 'm' terms and used that to prove it also works for 'm+1' terms. This shows the formula is correct for summing up all terms in this series.

🎯 Exam Tip: When dealing with complex series, verify the \( n^{th} \) term by substituting small values of n to ensure it correctly generates the terms of the series. This confirms you're working with the right statement P(n).

Question 16. Prove by the method of mathematical induction that \( \frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\frac{1}{10 \cdot 13}+\ldots+\frac{1}{(3 n+1)(3 n+4)}=\frac{n}{4(3 n+4)} \), for all natural numbers \( n \).
Answer: Let P(n) be the statement: \( \frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\frac{1}{10 \cdot 13}+\ldots+\frac{1}{(3 n+1)(3 n+4)}=\frac{n}{4(3 n+4)} \). This formula sums a series of fractions with products of numbers from an arithmetic progression in the denominator.
Step 1: Base Case (n=1)
We check if the statement is true for \( n=1 \).
LHS = \( \frac{1}{(3 \cdot 1+1)(3 \cdot 1+4)} = \frac{1}{4 \cdot 7} = \frac{1}{28} \)
RHS = \( \frac{1}{4(3 \cdot 1+4)} = \frac{1}{4 \cdot 7} = \frac{1}{28} \)
Since LHS = RHS, the statement P(1) is true.
Step 2: Inductive Hypothesis
We assume that the statement P(m) is true for some positive integer m. This means:
\( \frac{1}{4 \cdot 7}+\frac{1}{7 \cdot 10}+\ldots+\frac{1}{(3m+1)(3m+4)}=\frac{m}{4(3m+4)} \) (This is our Equation (1))
Step 3: Inductive Step
Now, we need to prove that P(m+1) is also true. This means we need to show that:
\( \frac{1}{4 \cdot 7}+\ldots+\frac{1}{(3m+1)(3m+4)} + \frac{1}{(3(m+1)+1)(3(m+1)+4)} = \frac{m+1}{4(3(m+1)+4)} \)
Which simplifies to:
\( \frac{1}{4 \cdot 7}+\ldots+\frac{1}{(3m+1)(3m+4)} + \frac{1}{(3m+4)(3m+7)} = \frac{m+1}{4(3m+7)} \)
Let's start with the LHS of P(m+1):
LHS = \( \left( \frac{1}{4 \cdot 7}+\ldots+\frac{1}{(3m+1)(3m+4)} \right) + \frac{1}{(3m+4)(3m+7)} \)
Using our assumption from Equation (1):
LHS = \( \frac{m}{4(3m+4)} + \frac{1}{(3m+4)(3m+7)} \)
To combine these fractions, find a common denominator, which is \( 4(3m+4)(3m+7) \):
LHS = \( \frac{m(3m+7) + 4}{4(3m+4)(3m+7)} \)
Expand the numerator:
LHS = \( \frac{3m^2+7m+4}{4(3m+4)(3m+7)} \)
Now, factor the quadratic expression \( 3m^2+7m+4 \). We look for two numbers that multiply to \( 3 \times 4 = 12 \) and add up to 7, which are 3 and 4.
\( 3m^2+7m+4 = 3m^2+3m+4m+4 = 3m(m+1) + 4(m+1) = (3m+4)(m+1) \)
So, substitute this back into the LHS:
LHS = \( \frac{(3m+4)(m+1)}{4(3m+4)(3m+7)} \)
Cancel out the common term \( (3m+4) \), assuming \( 3m+4 \neq 0 \) (which is true for natural numbers m):
LHS = \( \frac{m+1}{4(3m+7)} \)
This is exactly the RHS of P(m+1). So, P(m+1) is true.
Conclusion:
Since P(1) is true and P(m) implies P(m+1), by the principle of mathematical induction, the statement P(n) is true for all natural numbers \( n \).
In simple words: We checked if the formula works for the first term, and it did. Then we assumed it works for 'm' terms and used that to show it also works for 'm+1' terms. This means the formula is correct for adding up all terms in this sequence of fractions.

🎯 Exam Tip: Always look for opportunities to factorize and cancel common terms after combining fractions in the inductive step. This is a common pattern in these types of problems and simplifies the expression significantly.

Question 17. Let \( S(k) = 1 + 3 + 5 + ... + (2k - 1) = 3 + k^2 \), then which of the following is true?
(a) \( S(k) = S(k + 1) \)
(b) \( S(k) \neq S(k + 1) \)
(c) \( S(1) \) is correct
(d) Principle of mathematical induction can be used to prove the formula.
Answer: (b) \( S(k) \neq S(k + 1) \)
The given statement is \( S(k): 1 + 3 + 5 + ... + (2k - 1) = 3 + k^2 \).
We know that the sum of the first k odd numbers, \( 1 + 3 + 5 + ... + (2k - 1) \), is actually \( k^2 \).
So, the given statement \( S(k) \) means \( k^2 = 3 + k^2 \). This equation is false for all values of k.
Let's check each option:
(a) \( S(k) = S(k+1) \): If this were true, it would mean \( k^2 = (k+1)^2 \), which is false for all natural numbers k (only true for \( k = -1/2 \)). So, (a) is false.
(b) \( S(k) \neq S(k+1) \): This means that the sum of the first k odd numbers is not equal to the sum of the first (k+1) odd numbers. Since \( k^2 \neq (k+1)^2 \) for all natural numbers k, this statement is true.
(c) \( S(1) \) is correct: For \( k=1 \), the sum of the first odd number is \( 1 \). According to the given formula, \( S(1) = 3 + 1^2 = 3 + 1 = 4 \). Since \( 1 \neq 4 \), the statement \( S(1) \) is not correct. So, (c) is false.
(d) Principle of mathematical induction can be used to prove the formula: For mathematical induction to apply, the base case (P(1) or S(1) being true) must hold. Since \( S(1) \) is not correct, the principle of mathematical induction cannot be used to prove this formula. So, (d) is false.
Therefore, the only true option is (b).
In simple words: The formula given for S(k) is wrong because the sum of odd numbers is k-squared, not 3 plus k-squared. Because the formula is wrong from the start (for k=1), we cannot use induction to prove it. Also, the sum for 'k' numbers is never the same as for 'k+1' numbers.

🎯 Exam Tip: Be critical when a question defines a statement like \( S(k) \) with an equality that might be inherently false. First, verify the base case \( S(1) \). If the base case fails, then mathematical induction cannot be applied, and other options need to be evaluated based on the inherent truth of the mathematical expressions.