Samacheer Kalvi Class 11 Business Maths Solutions Chapter 2 Algebra Exercise 2.5

Samacheer Kalvi 11th Business Maths Algebra Ex 2.5 Text Book Back Questions and Answers

By the principle of mathematical induction, prove the following:

Question 1. \( 1^3 + 2^3 + ...... + n^3 = \frac{n^{2}(n+1)^{2}}{4} \) for all \( n \in N \).
Answer:
Let \( P(n) \) be the statement: \( 1^3 + 2^3 + ...... + n^3 = \frac{n^{2}(n+1)^{2}}{4} \) for all \( n \in N \).
**Step 1: Base Case (n = 1)**
LHS \( = 1^3 = 1 \)
RHS \( = \frac{1^{2}(1+1)^{2}}{4} = \frac{1 \times 2^{2}}{4} = \frac{4}{4} = 1 \)
Since LHS = RHS, \( P(1) \) is true.
**Step 2: Inductive Hypothesis**
Assume that \( P(n) \) is true for some positive integer \( k \).
So, \( P(k): 1^3 + 2^3 + ...... + k^3 = \frac{k^{2}(k+1)^{2}}{4} \)
**Step 3: Inductive Step**
We need to prove that \( P(k+1) \) is true. This means we need to show:
\( 1^3 + 2^3 + ...... + k^3 + (k+1)^3 = \frac{(k+1)^{2}((k+1)+1)^{2}}{4} = \frac{(k+1)^{2}(k+2)^{2}}{4} \)
Let's consider the LHS of \( P(k+1) \):
\( 1^3 + 2^3 + ...... + k^3 + (k+1)^3 \)
From our inductive hypothesis, we can replace \( 1^3 + 2^3 + ...... + k^3 \) with \( \frac{k^{2}(k+1)^{2}}{4} \).
\( = \frac{k^{2}(k+1)^{2}}{4} + (k+1)^3 \)
Now, we can take \( (k+1)^2 \) as a common factor:
\( = (k+1)^2 \left[ \frac{k^2}{4} + (k+1) \right] \)
To combine the terms inside the bracket, find a common denominator:
\( = (k+1)^2 \left[ \frac{k^2 + 4(k+1)}{4} \right] \)
Expand the numerator:
\( = (k+1)^2 \left[ \frac{k^2 + 4k + 4}{4} \right] \)
The expression \( k^2 + 4k + 4 \) is a perfect square, which is \( (k+2)^2 \).
\( = (k+1)^2 \left[ \frac{(k+2)^2}{4} \right] \)
\( = \frac{(k+1)^2(k+2)^2}{4} \)
This is the RHS of \( P(k+1) \).

\( \implies \) Since \( P(k+1) \) is true whenever \( P(k) \) is true, by the principle of mathematical induction, \( P(n) \) is true for all natural numbers \( n \).
In simple words: We checked if the formula works for the first number, \( n=1 \). Then we assumed it works for any number \( k \). Finally, we showed that if it works for \( k \), it must also work for the next number, \( k+1 \). Because of these steps, we know the formula works for all counting numbers.

🎯 Exam Tip: Remember to clearly state the base case, inductive hypothesis, and inductive step. Pay close attention to algebraic manipulations, especially factoring and finding common denominators, as these are crucial for connecting \( P(k) \) to \( P(k+1) \).

Question 2. \( 1.2 + 2.3 + 3.4 + ...... + n(n + 1) = \frac{n(n+1)(n+2)}{3} \), for all \( n \in N \).
Answer:
Let \( P(n) \) be the statement: \( 1.2 + 2.3 + 3.4 + ...... + n(n + 1) = \frac{n(n+1)(n+2)}{3} \) for all \( n \in N \).
**Step 1: Base Case (n = 1)**
LHS \( = 1(1+1) = 1(2) = 2 \)
RHS \( = \frac{1(1+1)(1+2)}{3} = \frac{1(2)(3)}{3} = \frac{6}{3} = 2 \)
Since LHS = RHS, \( P(1) \) is true.
**Step 2: Inductive Hypothesis**
Assume that \( P(n) \) is true for some positive integer \( k \).
So, \( P(k): 1.2 + 2.3 + 3.4 + ...... + k(k + 1) = \frac{k(k+1)(k+2)}{3} \)
**Step 3: Inductive Step**
We need to prove that \( P(k+1) \) is true. This means we need to show:
\( 1.2 + 2.3 + 3.4 + ...... + k(k+1) + (k+1)(k+2) = \frac{(k+1)((k+1)+1)((k+1)+2)}{3} = \frac{(k+1)(k+2)(k+3)}{3} \)
Let's consider the LHS of \( P(k+1) \):
\( [1.2 + 2.3 + 3.4 + ...... + k(k+1)] + (k+1)(k+2) \)
From our inductive hypothesis, we can replace the terms in the bracket:
\( = \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) \)
Now, take \( (k+1)(k+2) \) as a common factor:
\( = (k+1)(k+2) \left[ \frac{k}{3} + 1 \right] \)
Find a common denominator for the terms inside the bracket:
\( = (k+1)(k+2) \left[ \frac{k+3}{3} \right] \)
\( = \frac{(k+1)(k+2)(k+3)}{3} \)
This is the RHS of \( P(k+1) \).

\( \implies \) Since \( P(k+1) \) is true whenever \( P(k) \) is true, by the principle of mathematical induction, \( P(n) \) is true for all natural numbers \( n \).
In simple words: We first checked if the given rule works for \( n=1 \). Then, we assumed it works for any number \( k \). After that, we showed that if it works for \( k \), it will definitely work for the next number, \( k+1 \). This process shows the rule is correct for all natural numbers.

🎯 Exam Tip: For series questions, make sure to correctly identify the \( (k+1)^{th} \) term and factor it out from the expression to simplify the algebra. Remember to keep the common factor from the previous term of the sum.

Question 3. \( 4 + 8 + 12 + ....... + 4n = 2n(n + 1) \), for all \( n \in N \).
Answer:
Let \( P(n) \) be the statement: \( 4 + 8 + 12 + ....... + 4n = 2n(n + 1) \) for all \( n \in N \).
**Step 1: Base Case (n = 1)**
LHS \( = 4(1) = 4 \)
RHS \( = 2(1)(1 + 1) = 2(1)(2) = 4 \)
Since LHS = RHS, \( P(1) \) is true.
**Step 2: Inductive Hypothesis**
Assume that \( P(n) \) is true for some positive integer \( k \).
So, \( P(k): 4 + 8 + 12 + ....... + 4k = 2k(k + 1) \)
**Step 3: Inductive Step**
We need to prove that \( P(k+1) \) is true. This means we need to show:
\( 4 + 8 + 12 + ....... + 4k + 4(k+1) = 2(k+1)((k+1)+1) = 2(k+1)(k+2) \)
Let's consider the LHS of \( P(k+1) \):
\( [4 + 8 + 12 + ....... + 4k] + 4(k+1) \)
From our inductive hypothesis, we can replace the terms in the bracket:
\( = 2k(k + 1) + 4(k + 1) \)
Now, take \( (k+1) \) as a common factor:
\( = (k+1)[2k + 4] \)
Factor out 2 from the bracket:
\( = (k+1)2[k + 2] \)
\( = 2(k+1)(k+2) \)
This is the RHS of \( P(k+1) \).

\( \implies \) Since \( P(k+1) \) is true whenever \( P(k) \) is true, by the principle of mathematical induction, \( P(n) \) is true for all natural numbers \( n \).
In simple words: We checked the rule for \( n=1 \) and found it to be correct. Then we assumed the rule works for some number \( k \). Using that assumption, we proved that the rule must also work for the next number, \( k+1 \). This shows the rule is valid for all positive whole numbers.

🎯 Exam Tip: When proving divisibility or sum formulas by induction, always look for common factors after substituting the inductive hypothesis to simplify the expression and reach the \( P(k+1) \) form.

Question 4. \( 1 + 4 + 7 + ....... + (3n – 2) = \frac{n(3 n-1)}{2} \) for all \( n \in N \).
Answer:
Let \( P(n) \) be the statement: \( 1 + 4 + 7 + ....... + (3n – 2) = \frac{n(3 n-1)}{2} \) for all \( n \in N \).
**Step 1: Base Case (n = 1)**
LHS \( = 3(1) - 2 = 1 \)
RHS \( = \frac{1(3(1)-1)}{2} = \frac{1(3-1)}{2} = \frac{1(2)}{2} = 1 \)
Since LHS = RHS, \( P(1) \) is true.
**Step 2: Inductive Hypothesis**
Assume that \( P(n) \) is true for some positive integer \( k \).
So, \( P(k): 1 + 4 + 7 + ....... + (3k – 2) = \frac{k(3k-1)}{2} \)
**Step 3: Inductive Step**
We need to prove that \( P(k+1) \) is true. This means we need to show:
\( 1 + 4 + 7 + ....... + (3k – 2) + (3(k+1) – 2) = \frac{(k+1)(3(k+1)-1)}{2} = \frac{(k+1)(3k+2)}{2} \)
Let's consider the LHS of \( P(k+1) \):
\( [1 + 4 + 7 + ....... + (3k – 2)] + (3(k+1) – 2) \)
From our inductive hypothesis, we can replace the terms in the bracket:
\( = \frac{k(3k-1)}{2} + (3k + 3 – 2) \)
\( = \frac{k(3k-1)}{2} + (3k + 1) \)
Find a common denominator:
\( = \frac{k(3k-1) + 2(3k+1)}{2} \)
Expand the numerator:
\( = \frac{3k^2 - k + 6k + 2}{2} \)
Combine like terms:
\( = \frac{3k^2 + 5k + 2}{2} \)
Factor the quadratic expression in the numerator: \( 3k^2 + 5k + 2 = (3k+2)(k+1) \).
\( = \frac{(k+1)(3k+2)}{2} \)
This is the RHS of \( P(k+1) \).

\( \implies \) Since \( P(k+1) \) is true whenever \( P(k) \) is true, by the principle of mathematical induction, \( P(n) \) is true for all natural numbers \( n \).
In simple words: We confirmed the rule works for \( n=1 \). Then we supposed it works for any whole number \( k \). Based on this, we proved it must also work for \( k+1 \). This means the rule is correct for all positive whole numbers.

🎯 Exam Tip: When dealing with arithmetic series, correctly identifying the \( k+1 \) term and performing careful algebraic simplification of the numerator after finding a common denominator are key steps to success.

Question 5. \( 3^{2n} – 1 \) is divisible by 8, for all \( n \in N \).
Answer:
Let \( P(n) \) be the statement: \( 3^{2n} – 1 \) is divisible by 8 for all \( n \in N \).
**Step 1: Base Case (n = 1)**
For \( n=1 \), \( P(1) \) is \( 3^{2(1)} – 1 = 3^2 – 1 = 9 – 1 = 8 \).
Since 8 is divisible by 8, \( P(1) \) is true.
**Step 2: Inductive Hypothesis**
Assume that \( P(n) \) is true for some positive integer \( k \).
So, \( P(k): 3^{2k} – 1 \) is divisible by 8. This means \( 3^{2k} – 1 = 8m \) for some integer \( m \).
From this, we can write \( 3^{2k} = 8m + 1 \).
**Step 3: Inductive Step**
We need to prove that \( P(k+1) \) is true. This means we need to show:
\( 3^{2(k+1)} – 1 \) is divisible by 8.
Consider \( 3^{2(k+1)} – 1 \):
\( = 3^{2k+2} – 1 \)
\( = 3^{2k} \cdot 3^2 – 1 \)
\( = 3^{2k} \cdot 9 – 1 \)
Now, rewrite 9 as \( (8+1) \):
\( = 3^{2k}(8+1) – 1 \)
Distribute \( 3^{2k} \):
\( = 3^{2k} \cdot 8 + 3^{2k} \cdot 1 – 1 \)
\( = 3^{2k} \cdot 8 + (3^{2k} – 1) \)
Using our inductive hypothesis, we know \( 3^{2k} – 1 = 8m \):
\( = 3^{2k} \cdot 8 + 8m \)
Factor out 8:
\( = 8(3^{2k} + m) \)
This expression is clearly divisible by 8.

\( \implies \) Since \( P(k+1) \) is true whenever \( P(k) \) is true, by the principle of mathematical induction, \( P(n) \) is true for all natural numbers \( n \).
In simple words: We proved the statement works for \( n=1 \). Then, we assumed it works for a number \( k \). Using that assumption, we were able to show that it must also work for \( k+1 \). This confirms the statement is true for all positive whole numbers.

🎯 Exam Tip: For divisibility proofs, the key is to correctly substitute the inductive hypothesis (e.g., \( 3^{2k} = 8m+1 \)) into the \( P(k+1) \) expression and then factor out the divisor (e.g., 8).

Question 6. \( a^n – b^n \) is divisible by \( a – b \), for all \( n \in N \).
Answer:
Let \( P(n) \) be the statement: \( a^n – b^n \) is divisible by \( a – b \) for all \( n \in N \).
**Step 1: Base Case (n = 1)**
For \( n=1 \), \( P(1) \) is \( a^1 – b^1 = a – b \).
Since \( a-b \) is divisible by \( a-b \), \( P(1) \) is true.
**Step 2: Inductive Hypothesis**
Assume that \( P(n) \) is true for some positive integer \( k \).
So, \( P(k): a^k – b^k \) is divisible by \( a – b \). This means \( a^k – b^k = m(a – b) \) for some integer \( m \).
From this, we can write \( a^k = b^k + m(a – b) \).
**Step 3: Inductive Step**
We need to prove that \( P(k+1) \) is true. This means we need to show:
\( a^{k+1} – b^{k+1} \) is divisible by \( a – b \).
Consider \( a^{k+1} – b^{k+1} \):
\( = a^k \cdot a – b^k \cdot b \)
Substitute \( a^k = b^k + m(a – b) \) into the expression:
\( = [b^k + m(a – b)]a – b^k \cdot b \)
Distribute \( a \):
\( = b^k \cdot a + am(a – b) – b^k \cdot b \)
Rearrange the terms:
\( = b^k(a – b) + am(a – b) \)
Now, factor out \( (a – b) \):
\( = (a – b)(b^k + am) \)
This expression is clearly divisible by \( a – b \).

\( \implies \) Since \( P(k+1) \) is true whenever \( P(k) \) is true, by the principle of mathematical induction, \( P(n) \) is true for all natural numbers \( n \).
In simple words: First, we checked that the rule holds for \( n=1 \). Then we assumed it works for any whole number \( k \). By using this assumption, we showed that the rule must also be true for the next number, \( k+1 \). This proves the rule works for all positive whole numbers.

🎯 Exam Tip: In divisibility proofs, the key step is often substituting the inductive hypothesis (e.g., \( a^k = b^k + m(a-b) \)) and then carefully manipulating the expression to factor out the desired divisor.

Question 7. \( 5^{2n} – 1 \) is divisible by 24, for all \( n \in N \).
Answer:
Let \( P(n) \) be the statement: \( 5^{2n} – 1 \) is divisible by 24 for all \( n \in N \).
**Step 1: Base Case (n = 1)**
For \( n=1 \), \( P(1) \) is \( 5^{2(1)} – 1 = 5^2 – 1 = 25 – 1 = 24 \).
Since 24 is divisible by 24, \( P(1) \) is true.
**Step 2: Inductive Hypothesis**
Assume that \( P(n) \) is true for some positive integer \( k \).
So, \( P(k): 5^{2k} – 1 \) is divisible by 24. This means \( 5^{2k} – 1 = 24m \) for some integer \( m \).
From this, we can write \( 5^{2k} = 24m + 1 \).
**Step 3: Inductive Step**
We need to prove that \( P(k+1) \) is true. This means we need to show:
\( 5^{2(k+1)} – 1 \) is divisible by 24.
Consider \( 5^{2(k+1)} – 1 \):
\( = 5^{2k+2} – 1 \)
\( = 5^{2k} \cdot 5^2 – 1 \)
\( = 5^{2k} \cdot 25 – 1 \)
Now, rewrite 25 as \( (24+1) \):
\( = 5^{2k}(24+1) – 1 \)
Distribute \( 5^{2k} \):
\( = 5^{2k} \cdot 24 + 5^{2k} \cdot 1 – 1 \)
\( = 24 \cdot 5^{2k} + (5^{2k} – 1) \)
Using our inductive hypothesis, we know \( 5^{2k} – 1 = 24m \):
\( = 24 \cdot 5^{2k} + 24m \)
Factor out 24:
\( = 24(5^{2k} + m) \)
This expression is clearly divisible by 24.

\( \implies \) Since \( P(k+1) \) is true whenever \( P(k) \) is true, by the principle of mathematical induction, \( P(n) \) is true for all natural numbers \( n \).
In simple words: We checked the statement for \( n=1 \) and it was true. Then we assumed it is true for any number \( k \). By using this assumption, we showed that the statement must also be true for \( k+1 \). This means the statement holds for all positive whole numbers.

🎯 Exam Tip: When proving divisibility statements, look for opportunities to split a constant (like 25 into \( 24+1 \)) to create terms that allow you to substitute the inductive hypothesis and factor out the divisor.

Question 8. \( n(n + 1) (n + 2) \) is divisible by 6, for all \( n \in N \).
Answer:
Let \( P(n) \) be the statement: \( n(n + 1) (n + 2) \) is divisible by 6 for all \( n \in N \).
**Step 1: Base Case (n = 1)**
For \( n=1 \), \( P(1) \) is \( 1(1+1)(1+2) = 1(2)(3) = 6 \).
Since 6 is divisible by 6, \( P(1) \) is true.
**Step 2: Inductive Hypothesis**
Assume that \( P(n) \) is true for some positive integer \( k \).
So, \( P(k): k(k + 1) (k + 2) \) is divisible by 6. This means \( k(k + 1) (k + 2) = 6m \) for some integer \( m \).
**Step 3: Inductive Step**
We need to prove that \( P(k+1) \) is true. This means we need to show:
\( (k+1)((k+1)+1)((k+1)+2) = (k+1)(k+2)(k+3) \) is divisible by 6.
Consider \( (k+1)(k+2)(k+3) \):
We can split the term \( (k+3) \) as \( (k+1)(k+2)(k+3) = (k+1)(k+2)(k + 3) \)
\( = (k+1)(k+2) \cdot k + (k+1)(k+2) \cdot 3 \)
\( = k(k+1)(k+2) + 3(k+1)(k+2) \)
Using our inductive hypothesis, we know \( k(k+1)(k+2) = 6m \):
\( = 6m + 3(k+1)(k+2) \)
Now, let's look at the second term, \( 3(k+1)(k+2) \).
The product \( (k+1)(k+2) \) represents the product of two consecutive integers. One of these integers must always be an even number. So, \( (k+1)(k+2) \) is always divisible by 2. This means \( (k+1)(k+2) = 2j \) for some integer \( j \).
Therefore, \( 3(k+1)(k+2) = 3(2j) = 6j \). This means \( 3(k+1)(k+2) \) is divisible by 6.
Since both \( 6m \) and \( 3(k+1)(k+2) \) (which equals \( 6j \)) are divisible by 6, their sum is also divisible by 6.
\( = 6m + 6j = 6(m+j) \)
This expression is clearly divisible by 6.

\( \implies \) Since \( P(k+1) \) is true whenever \( P(k) \) is true, by the principle of mathematical induction, \( P(n) \) is true for all natural numbers \( n \).
In simple words: We confirmed the statement works for \( n=1 \). Then we assumed it is true for any number \( k \). Using this, we showed it must also be true for \( k+1 \). This proves the statement is correct for all positive whole numbers. An important part is knowing that out of any two numbers in a row, one must be even.

🎯 Exam Tip: When proving divisibility by a composite number (like 6), remember to show divisibility by its prime factors (like 2 and 3). For consecutive integers, one is always even; for three consecutive integers, one is always a multiple of 3.

Question 9. \( 2^n > n \), for all \( n \in N \).
Answer:
Let \( P(n) \) be the statement: \( 2^n > n \) for all \( n \in N \).
**Step 1: Base Case (n = 1)**
For \( n=1 \), \( P(1) \) is \( 2^1 > 1 \), which means \( 2 > 1 \).
This statement is true.
**Step 2: Inductive Hypothesis**
Assume that \( P(n) \) is true for some positive integer \( k \).
So, \( P(k): 2^k > k \) for \( k \ge 1 \).
**Step 3: Inductive Step**
We need to prove that \( P(k+1) \) is true. This means we need to show:
\( 2^{k+1} > k+1 \).
From our inductive hypothesis, we know \( 2^k > k \).
Multiply both sides of the inequality by 2:
\( 2 \cdot 2^k > 2 \cdot k \)
\( \implies 2^{k+1} > 2k \)
Now, we need to show that \( 2k \ge k+1 \) for \( k \ge 1 \).
We can rewrite \( 2k \) as \( k+k \).
Since \( k \ge 1 \), it means \( k \) is at least 1. So, \( k \ge 1 \).
Therefore, \( k+k \ge k+1 \).
This means \( 2k \ge k+1 \).
Combining the inequalities:
\( 2^{k+1} > 2k \)

\( \implies 2^{k+1} \ge k+1 \)
Therefore, \( 2^{k+1} > k+1 \) is true.

\( \implies \) Since \( P(k+1) \) is true whenever \( P(k) \) is true, by the principle of mathematical induction, \( P(n) \) is true for all natural numbers \( n \).
In simple words: We showed that the rule works for \( n=1 \). Then, we assumed it works for a number \( k \). With this assumption, we proved that the rule must also work for \( k+1 \). This means that for any positive whole number, 2 raised to that number will always be bigger than the number itself.

🎯 Exam Tip: For inequality proofs, remember to use the inductive hypothesis effectively. If you have \( 2^{k+1} > 2k \) and you need to show \( 2^{k+1} > k+1 \), prove that \( 2k \ge k+1 \) separately, or demonstrate that \( 2k \) is a value greater than \( k+1 \).