Samacheer Kalvi Class 11 Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Exercise 4.4

Question 1. By the principle of mathematical induction, prove that, for n ≥ 1 \( 1^3 + 2^3 + 3^3 + ... + n^3 = \left(\frac{n(n+1)}{2}\right)^{2} \)
Answer: Let the given statement be \( P(n) \).
**Step 1: Base Case (Verify for \( n=1 \))**
LHS: \( P(1) = 1^3 = 1 \)
RHS: \( \left(\frac{1(1+1)}{2}\right)^{2} = \left(\frac{1 \cdot 2}{2}\right)^{2} = 1^2 = 1 \)
Since LHS = RHS, \( P(1) \) is true.
**Step 2: Inductive Hypothesis (Assume for \( n=k \))**
Assume that the statement is true for some positive integer \( k \geq 1 \).
So, \( P(k) : 1^3 + 2^3 + 3^3 + ... + k^3 = \left(\frac{k(k+1)}{2}\right)^{2} \) is true.
**Step 3: Inductive Step (Prove for \( n=k+1 \))**
We need to prove that \( P(k+1) \) is true, i.e., \( 1^3 + 2^3 + ... + k^3 + (k+1)^3 = \left(\frac{(k+1)((k+1)+1)}{2}\right)^{2} = \left(\frac{(k+1)(k+2)}{2}\right)^{2} \).
Start with the LHS of \( P(k+1) \):
\( LHS = (1^3 + 2^3 + ... + k^3) + (k+1)^3 \)
Substitute \( P(k) \) from the inductive hypothesis:
\( LHS = \left(\frac{k(k+1)}{2}\right)^{2} + (k+1)^3 \)
\( = \frac{k^2(k+1)^2}{4} + (k+1)^3 \)
Find a common denominator:
\( = \frac{k^2(k+1)^2 + 4(k+1)^3}{4} \)
Factor out \( (k+1)^2 \) from the numerator:
\( = \frac{(k+1)^2 [k^2 + 4(k+1)]}{4} \)
Simplify the expression inside the square bracket:
\( = \frac{(k+1)^2 [k^2 + 4k + 4]}{4} \)
Recognize \( k^2 + 4k + 4 \) as a perfect square \( (k+2)^2 \):
\( = \frac{(k+1)^2 (k+2)^2}{4} \)
Group the terms to match the RHS form:
\( = \left(\frac{(k+1)(k+2)}{2}\right)^{2} = RHS \)
Thus, \( P(k+1) \) is true whenever \( P(k) \) is true.
**Conclusion:** By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \geq 1 \).
In simple words: We check if the statement works for the first number, \( n=1 \). Then, we assume it works for any number \( k \). Using that assumption, we show it also works for the next number, \( k+1 \). Since it works for 1, and for \( k+1 \) if it works for \( k \), it must be true for all numbers.

🎯 Exam Tip: Clearly state the three steps: Base case, Inductive hypothesis, and Inductive step. Show all algebraic manipulations to reach the \( n=k+1 \) form. This structured approach helps ensure no steps are missed and full marks are awarded.

Question 2. By the principle of mathematical induction, prove that, for n ≥ 1 \( 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = \frac{n(2n - 1)(2n + 1)}{3} \)
Answer: Let the given statement be \( P(n) \).
**Step 1: Base Case (Verify for \( n=1 \))**
LHS: \( P(1) = 1^2 = 1 \)
RHS: \( \frac{1(2 \cdot 1 - 1)(2 \cdot 1 + 1)}{3} = \frac{1(1)(3)}{3} = 1 \)
Since LHS = RHS, \( P(1) \) is true.
**Step 2: Inductive Hypothesis (Assume for \( n=k \))**
Assume that the statement is true for some positive integer \( k \geq 1 \).
So, \( P(k) : 1^2 + 3^2 + 5^2 + ... + (2k-1)^2 = \frac{k(2k-1)(2k+1)}{3} \) is true.
**Step 3: Inductive Step (Prove for \( n=k+1 \))**
We need to prove that \( P(k+1) \) is true, i.e., \( 1^2 + 3^2 + ... + (2k-1)^2 + (2(k+1)-1)^2 = \frac{(k+1)(2(k+1)-1)(2(k+1)+1)}{3} = \frac{(k+1)(2k+1)(2k+3)}{3} \).
Start with the LHS of \( P(k+1) \):
\( LHS = (1^2 + 3^2 + ... + (2k-1)^2) + (2(k+1)-1)^2 \)
\( = (1^2 + 3^2 + ... + (2k-1)^2) + (2k+1)^2 \)
Substitute \( P(k) \) from the inductive hypothesis:
\( LHS = \frac{k(2k-1)(2k+1)}{3} + (2k+1)^2 \)
Factor out \( (2k+1) \) from the expression:
\( = (2k+1) \left[ \frac{k(2k-1)}{3} + (2k+1) \right] \)
Combine terms inside the square bracket by finding a common denominator:
\( = (2k+1) \left[ \frac{k(2k-1) + 3(2k+1)}{3} \right] \)
Expand and simplify the numerator:
\( = (2k+1) \left[ \frac{2k^2 - k + 6k + 3}{3} \right] \)
\( = (2k+1) \left[ \frac{2k^2 + 5k + 3}{3} \right] \)
Factor the quadratic \( 2k^2 + 5k + 3 \). It can be factored as \( (k+1)(2k+3) \):
\( = \frac{(2k+1)(k+1)(2k+3)}{3} \)
Rearrange the terms to match the RHS:
\( = \frac{(k+1)(2k+1)(2k+3)}{3} = RHS \)
Thus, \( P(k+1) \) is true whenever \( P(k) \) is true.
**Conclusion:** By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \geq 1 \).
In simple words: First, we prove the statement is true for the number 1. Then, we assume it's true for any number \( k \). Using this assumption, we show it must also be true for the next number, \( k+1 \). This process shows the statement is true for all whole numbers starting from 1.

🎯 Exam Tip: When factoring a quadratic expression like \( 2k^2+5k+3 \), remember to look for factors of \( ac \) that add up to \( b \). This helps to simplify the expression correctly.

Question 3. Prove that the sum of the first 'n' non-zero even numbers is \( n^2 + n \).
Answer: Let the given statement be \( P(n) : 2 + 4 + 6 + ... + 2n = n^2 + n \).
**Step 1: Base Case (Verify for \( n=1 \))**
LHS: \( P(1) = 2 \)
RHS: \( 1^2 + 1 = 1 + 1 = 2 \)
Since LHS = RHS, \( P(1) \) is true.
**Step 2: Inductive Hypothesis (Assume for \( n=k \))**
Assume that the statement is true for some positive integer \( k \geq 1 \).
So, \( P(k) : 2 + 4 + 6 + ... + 2k = k^2 + k \) is true.
**Step 3: Inductive Step (Prove for \( n=k+1 \))**
We need to prove that \( P(k+1) \) is true, i.e., \( 2 + 4 + ... + 2k + 2(k+1) = (k+1)^2 + (k+1) \).
Start with the LHS of \( P(k+1) \):
\( LHS = (2 + 4 + 6 + ... + 2k) + 2(k+1) \)
Substitute \( P(k) \) from the inductive hypothesis:
\( LHS = (k^2 + k) + 2(k+1) \)
Expand and simplify the expression:
\( = k^2 + k + 2k + 2 \)
\( = k^2 + 3k + 2 \)
Factor the quadratic expression:
\( = (k+1)(k+2) \)
Rearrange the terms to match the RHS form \( n^2+n \) for \( n=k+1 \):
\( = (k+1)[(k+1)+1] = (k+1)^2 + (k+1) = RHS \)
Thus, \( P(k+1) \) is true whenever \( P(k) \) is true.
**Conclusion:** By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \geq 1 \).
In simple words: We check the rule for the first even number (2). Then we assume it holds for any 'k' even numbers. Using that, we prove it works for 'k+1' even numbers. This way, we show the rule holds for all natural numbers.

🎯 Exam Tip: Remember that the \( (k+1) \)-th even number is \( 2(k+1) \). Correctly identifying this next term is crucial for setting up the inductive step.

Question 4. By the principle of mathematical induction, prove that, for n ≥ 1, \( 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + ... + n(n + 1) = \frac{n(n+1)(n+2)}{3} \).
Answer: Let the given statement be \( P(n) \).
**Step 1: Base Case (Verify for \( n=1 \))**
LHS: \( P(1) = 1 \cdot 2 = 2 \)
RHS: \( \frac{1(1+1)(1+2)}{3} = \frac{1 \cdot 2 \cdot 3}{3} = 2 \)
Since LHS = RHS, \( P(1) \) is true.
**Step 2: Inductive Hypothesis (Assume for \( n=k \))**
Assume that the statement is true for some positive integer \( k \geq 1 \).
So, \( P(k) : 1 \cdot 2 + 2 \cdot 3 + ... + k(k+1) = \frac{k(k+1)(k+2)}{3} \) is true.
**Step 3: Inductive Step (Prove for \( n=k+1 \))**
We need to prove that \( P(k+1) \) is true, i.e., \( 1 \cdot 2 + ... + k(k+1) + (k+1)((k+1)+1) = \frac{(k+1)((k+1)+1)((k+1)+2)}{3} = \frac{(k+1)(k+2)(k+3)}{3} \).
Start with the LHS of \( P(k+1) \):
\( LHS = (1 \cdot 2 + ... + k(k+1)) + (k+1)(k+2) \)
Substitute \( P(k) \) from the inductive hypothesis:
\( LHS = \frac{k(k+1)(k+2)}{3} + (k+1)(k+2) \)
Factor out the common term \( (k+1)(k+2) \):
\( = (k+1)(k+2) \left[ \frac{k}{3} + 1 \right] \)
Combine terms inside the square bracket:
\( = (k+1)(k+2) \left[ \frac{k+3}{3} \right] \)
Rearrange the terms to match the RHS form:
\( = \frac{(k+1)(k+2)(k+3)}{3} = RHS \)
Thus, \( P(k+1) \) is true whenever \( P(k) \) is true.
**Conclusion:** By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \geq 1 \).
In simple words: First, we confirm the rule works for \( n=1 \). Then, we suppose it works for some number \( k \). Using that, we show it also works for the next number, \( k+1 \). This proves the rule is true for all natural numbers.

🎯 Exam Tip: Factoring out the common terms \( (k+1)(k+2) \) is key to simplifying the expression in the inductive step, making it easier to reach the target form.

Question 5. Using the Mathematical induction, show that for any natural number n ≥ 2, \( \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right) = \frac{n+1}{2n} \).
Answer: Let the given statement be \( P(n) \).
**Step 1: Base Case (Verify for \( n=2 \))**
LHS: \( P(2) = \left(1-\frac{1}{2^2}\right) = 1-\frac{1}{4} = \frac{3}{4} \)
RHS: \( \frac{2+1}{2 \cdot 2} = \frac{3}{4} \)
Since LHS = RHS, \( P(2) \) is true.
**Step 2: Inductive Hypothesis (Assume for \( n=k \))**
Assume that the statement is true for some positive integer \( k \geq 2 \).
So, \( P(k) : \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{k^2}\right) = \frac{k+1}{2k} \) is true.
**Step 3: Inductive Step (Prove for \( n=k+1 \))**
We need to prove that \( P(k+1) \) is true, i.e., \( \left(1-\frac{1}{2^2}\right)...\left(1-\frac{1}{k^2}\right)\left(1-\frac{1}{(k+1)^2}\right) = \frac{(k+1)+1}{2(k+1)} = \frac{k+2}{2(k+1)} \).
Start with the LHS of \( P(k+1) \):
\( LHS = \left[\left(1-\frac{1}{2^2}\right)...\left(1-\frac{1}{k^2}\right)\right] \left(1-\frac{1}{(k+1)^2}\right) \)
Substitute \( P(k) \) from the inductive hypothesis:
\( LHS = \frac{k+1}{2k} \left(1-\frac{1}{(k+1)^2}\right) \)
Simplify the second factor by finding a common denominator:
\( = \frac{k+1}{2k} \left(\frac{(k+1)^2-1}{(k+1)^2}\right) \)
Expand the numerator \( (k+1)^2-1 \):
\( = \frac{k+1}{2k} \left(\frac{k^2+2k+1-1}{(k+1)^2}\right) \)
\( = \frac{k+1}{2k} \left(\frac{k^2+2k}{(k+1)^2}\right) \)
Factor out \( k \) from the numerator of the second factor:
\( = \frac{k+1}{2k} \left(\frac{k(k+2)}{(k+1)^2}\right) \)
Cancel common terms \( (k+1) \) and \( k \):
\( = \frac{k(k+1)(k+2)}{2k(k+1)^2} \)
\( = \frac{k+2}{2(k+1)} = RHS \)
Thus, \( P(k+1) \) is true whenever \( P(k) \) is true.
**Conclusion:** By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \geq 2 \).
In simple words: We first check if the statement works for \( n=2 \). Then, we assume it works for some number \( k \). We use this to show it also works for \( k+1 \). This process confirms the statement is true for all natural numbers starting from 2.

🎯 Exam Tip: Remember to factor and simplify expressions carefully. Using the difference of squares identity \( a^2-b^2 = (a-b)(a+b) \) is helpful for terms like \( (k+1)^2-1 \).

Question 6. Using the Mathematical induction, show that for any natural number n ≥ 2, \( \frac{1}{1+2} + \frac{1}{1+2+3} + \frac{1}{1+2+3+4} + ... + \frac{1}{1+2+3+...+n} = \frac{n-1}{n+1} \).
Answer: Let the given statement be \( P(n) \).
The general term in the sum is \( \frac{1}{1+2+...+r} \). We know that the sum of the first \( r \) natural numbers is \( \frac{r(r+1)}{2} \).
So, the general term can be written as \( \frac{1}{r(r+1)/2} = \frac{2}{r(r+1)} \).
Thus, \( P(n) : \sum_{r=2}^{n} \frac{2}{r(r+1)} = \frac{n-1}{n+1} \).
**Step 1: Base Case (Verify for \( n=2 \))**
LHS: \( P(2) = \frac{1}{1+2} = \frac{1}{3} \)
RHS: \( \frac{2-1}{2+1} = \frac{1}{3} \)
Since LHS = RHS, \( P(2) \) is true.
**Step 2: Inductive Hypothesis (Assume for \( n=k \))**
Assume that the statement is true for some positive integer \( k \geq 2 \).
So, \( P(k) : \frac{1}{1+2} + ... + \frac{1}{1+2+...+k} = \frac{k-1}{k+1} \) is true.
**Step 3: Inductive Step (Prove for \( n=k+1 \))**
We need to prove that \( P(k+1) \) is true, i.e., \( \left(\frac{1}{1+2} + ... + \frac{1}{1+...+k}\right) + \frac{1}{1+...+k+(k+1)} = \frac{(k+1)-1}{(k+1)+1} = \frac{k}{k+2} \).
Start with the LHS of \( P(k+1) \):
\( LHS = \left(\frac{1}{1+2} + ... + \frac{1}{1+...+k}\right) + \frac{1}{1+...+k+(k+1)} \)
The last term is \( \frac{2}{(k+1)((k+1)+1)} = \frac{2}{(k+1)(k+2)} \).
Substitute \( P(k) \) from the inductive hypothesis:
\( LHS = \frac{k-1}{k+1} + \frac{2}{(k+1)(k+2)} \)
Find a common denominator \( (k+1)(k+2) \):
\( = \frac{(k-1)(k+2) + 2}{(k+1)(k+2)} \)
Expand and simplify the numerator:
\( = \frac{k^2+2k-k-2+2}{(k+1)(k+2)} \)
\( = \frac{k^2+k}{(k+1)(k+2)} \)
Factor out \( k \) from the numerator:
\( = \frac{k(k+1)}{(k+1)(k+2)} \)
Cancel the common term \( (k+1) \):
\( = \frac{k}{k+2} = RHS \)
Thus, \( P(k+1) \) is true whenever \( P(k) \) is true.
**Conclusion:** By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \geq 2 \).
In simple words: First, we check if the statement works for \( n=2 \). Then, we assume it holds for some \( k \). Using that, we show it must also hold for \( k+1 \). This proves the rule is true for all natural numbers starting from 2. The key is to simplify the sum of numbers in the denominator.

🎯 Exam Tip: Recognize that the denominator \( 1+2+...+r \) is the sum of an arithmetic progression, which simplifies to \( \frac{r(r+1)}{2} \). This transformation is essential for solving the problem efficiently.

Question 7. Using the mathematical induction, show that for any natural number n, \( \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + ... + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)} \).
Answer: Let the given statement be \( P(n) \).
**Step 1: Base Case (Verify for \( n=1 \))**
LHS: \( P(1) = \frac{1}{1 \cdot 2 \cdot 3} = \frac{1}{6} \)
RHS: \( \frac{1(1+3)}{4(1+1)(1+2)} = \frac{1 \cdot 4}{4 \cdot 2 \cdot 3} = \frac{4}{24} = \frac{1}{6} \)
Since LHS = RHS, \( P(1) \) is true.
**Step 2: Inductive Hypothesis (Assume for \( n=k \))**
Assume that the statement is true for some positive integer \( k \geq 1 \).
So, \( P(k) : \frac{1}{1 \cdot 2 \cdot 3} + ... + \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)} \) is true.
**Step 3: Inductive Step (Prove for \( n=k+1 \))**
We need to prove that \( P(k+1) \) is true, i.e., \( \left(\frac{1}{1 \cdot 2 \cdot 3} + ... + \frac{1}{k(k+1)(k+2)}\right) + \frac{1}{(k+1)((k+1)+1)((k+1)+2)} = \frac{(k+1)((k+1)+3)}{4((k+1)+1)((k+1)+2)} = \frac{(k+1)(k+4)}{4(k+2)(k+3)} \).
Start with the LHS of \( P(k+1) \):
\( LHS = \left(\frac{1}{1 \cdot 2 \cdot 3} + ... + \frac{1}{k(k+1)(k+2)}\right) + \frac{1}{(k+1)(k+2)(k+3)} \)
Substitute \( P(k) \) from the inductive hypothesis:
\( LHS = \frac{k(k+3)}{4(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)} \)
Find a common denominator \( 4(k+1)(k+2)(k+3) \):
\( = \frac{k(k+3)^2 + 4}{4(k+1)(k+2)(k+3)} \)
Expand and simplify the numerator:
\( = \frac{k(k^2+6k+9) + 4}{4(k+1)(k+2)(k+3)} \)
\( = \frac{k^3+6k^2+9k+4}{4(k+1)(k+2)(k+3)} \)
Factor the cubic numerator \( k^3+6k^2+9k+4 \). We can see that \( k=-1 \) is a root (\( (-1)^3+6(-1)^2+9(-1)+4 = -1+6-9+4=0 \)), so \( (k+1) \) is a factor. Dividing gives \( (k+1)(k^2+5k+4) \). Further factoring \( k^2+5k+4 = (k+1)(k+4) \).
So, the numerator is \( (k+1)(k+1)(k+4) = (k+1)^2(k+4) \).
Substitute the factored numerator back:
\( = \frac{(k+1)^2(k+4)}{4(k+1)(k+2)(k+3)} \)
Cancel one \( (k+1) \) term:
\( = \frac{(k+1)(k+4)}{4(k+2)(k+3)} = RHS \)
Thus, \( P(k+1) \) is true whenever \( P(k) \) is true.
**Conclusion:** By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \geq 1 \).
In simple words: We first check if the rule is true for \( n=1 \). Then, we assume it works for some number \( k \). Using that assumption, we show it also works for \( k+1 \). This method confirms the rule is true for all natural numbers. Finding factors for the top part of the fraction can be tricky but is important.

🎯 Exam Tip: When dealing with cubic polynomials in the numerator, try to find factors using the Rational Root Theorem (by testing simple integer roots like \(\pm 1, \pm 2\)) or by guessing factors based on the desired final form.

Question 8. Using the Mathematical induction, show that for any natural number n, \( \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + ... + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4} \).
Answer: Let the given statement be \( P(n) \).
**Step 1: Base Case (Verify for \( n=1 \))**
LHS: \( P(1) = \frac{1}{(3 \cdot 1-1)(3 \cdot 1+2)} = \frac{1}{(2)(5)} = \frac{1}{10} \)
RHS: \( \frac{1}{6 \cdot 1+4} = \frac{1}{6+4} = \frac{1}{10} \)
Since LHS = RHS, \( P(1) \) is true.
**Step 2: Inductive Hypothesis (Assume for \( n=k \))**
Assume that the statement is true for some positive integer \( k \geq 1 \).
So, \( P(k) : \frac{1}{2 \cdot 5} + ... + \frac{1}{(3k-1)(3k+2)} = \frac{k}{6k+4} \) is true.
**Step 3: Inductive Step (Prove for \( n=k+1 \))**
We need to prove that \( P(k+1) \) is true, i.e., \( \left(\frac{1}{2 \cdot 5} + ... + \frac{1}{(3k-1)(3k+2)}\right) + \frac{1}{(3(k+1)-1)(3(k+1)+2)} = \frac{k+1}{6(k+1)+4} = \frac{k+1}{6k+10} \).
Start with the LHS of \( P(k+1) \):
\( LHS = \left(\frac{1}{2 \cdot 5} + ... + \frac{1}{(3k-1)(3k+2)}\right) + \frac{1}{(3k+3-1)(3k+3+2)} \)
\( = \left(\frac{1}{2 \cdot 5} + ... + \frac{1}{(3k-1)(3k+2)}\right) + \frac{1}{(3k+2)(3k+5)} \)
Substitute \( P(k) \) from the inductive hypothesis:
\( LHS = \frac{k}{6k+4} + \frac{1}{(3k+2)(3k+5)} \)
Factor the denominator of the first term: \( 6k+4 = 2(3k+2) \).
\( = \frac{k}{2(3k+2)} + \frac{1}{(3k+2)(3k+5)} \)
Find a common denominator \( 2(3k+2)(3k+5) \):
\( = \frac{k(3k+5) + 2}{2(3k+2)(3k+5)} \)
Expand and simplify the numerator:
\( = \frac{3k^2+5k+2}{2(3k+2)(3k+5)} \)
Factor the quadratic numerator \( 3k^2+5k+2 \). It can be factored as \( (3k+2)(k+1) \):
\( = \frac{(3k+2)(k+1)}{2(3k+2)(3k+5)} \)
Cancel the common term \( (3k+2) \):
\( = \frac{k+1}{2(3k+5)} \)
Simplify the denominator to match the RHS form:
\( = \frac{k+1}{6k+10} = RHS \)
Thus, \( P(k+1) \) is true whenever \( P(k) \) is true.
**Conclusion:** By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \geq 1 \).
In simple words: We first check if the formula works for \( n=1 \). Then, we assume it's true for some number \( k \). Using this, we prove it must also be true for \( k+1 \). This shows the formula works for all natural numbers. The key is to find a common denominator and factor the top part of the fraction.

🎯 Exam Tip: Always factor the denominators and numerators as much as possible to identify common terms for cancellation. This makes algebraic manipulation much easier and helps avoid errors.

Question 8. Using the Mathematical induction, show that for any natural number n, \( \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4} \)
Answer: Let the given statement be \( P(n) \):
\( P(n) : \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4} \)
Step 1: Verify for \( n = 1 \)
For \( n = 1 \), the LHS is the first term: \( \frac{1}{2 \cdot 5} = \frac{1}{10} \)
The RHS is: \( \frac{1}{6(1)+4} = \frac{1}{6+4} = \frac{1}{10} \)
Since LHS = RHS, the statement \( P(1) \) is true.
Step 2: Assume for \( n = k \)
Assume that \( P(k) \) is true for some natural number \( k \ge 1 \).
So, \( \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3k-1)(3k+2)} = \frac{k}{6k+4} \)
Step 3: Prove for \( n = k+1 \)
We need to prove that \( P(k+1) \) is true. This means:
\( P(k+1) = \frac{1}{2 \cdot 5} + \frac{1}{5 \cdot 8} + \frac{1}{8 \cdot 11} + \dots + \frac{1}{(3k-1)(3k+2)} + \frac{1}{(3(k+1)-1)(3(k+1)+2)} \)
We can write this as:
\( P(k+1) = P(k) + \frac{1}{(3k+3-1)(3k+3+2)} \)
\( P(k+1) = P(k) + \frac{1}{(3k+2)(3k+5)} \)
Now, substitute \( P(k) \) from Step 2:
\( P(k+1) = \frac{k}{6k+4} + \frac{1}{(3k+2)(3k+5)} \)
\( P(k+1) = \frac{k}{2(3k+2)} + \frac{1}{(3k+2)(3k+5)} \)
To combine these, find a common denominator:
\( P(k+1) = \frac{k(3k+5)}{2(3k+2)(3k+5)} + \frac{2}{2(3k+2)(3k+5)} \)
\( P(k+1) = \frac{k(3k+5)+2}{2(3k+2)(3k+5)} \)
\( P(k+1) = \frac{3k^2+5k+2}{2(3k+2)(3k+5)} \)
Factorize the numerator \( 3k^2+5k+2 \). We can find factors by looking for two numbers that multiply to \( 3 \times 2 = 6 \) and add to 5 (which are 2 and 3).
\( 3k^2+5k+2 = 3k^2+3k+2k+2 = 3k(k+1)+2(k+1) = (3k+2)(k+1) \)
Substitute this back into the expression for \( P(k+1) \):
\( P(k+1) = \frac{(3k+2)(k+1)}{2(3k+2)(3k+5)} \)
Cancel out the common term \( (3k+2) \):
\( P(k+1) = \frac{k+1}{2(3k+5)} \)
Now, we want to show this matches the RHS for \( n=k+1 \), which is \( \frac{k+1}{6(k+1)+4} \).
Let's simplify the denominator of our result: \( 2(3k+5) = 6k+10 \).
So, \( P(k+1) = \frac{k+1}{6k+10} \)
The target RHS for \( n=k+1 \) is \( \frac{k+1}{6(k+1)+4} = \frac{k+1}{6k+6+4} = \frac{k+1}{6k+10} \)
Since the calculated \( P(k+1) \) matches the RHS for \( n=k+1 \), the statement \( P(k+1) \) is true.
By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \). This means the given sum formula is correct for all natural numbers.
In simple words: We used a special proof method called induction. First, we checked if the formula works for the very first number, 1. Then, we assumed it works for any number 'k'. Finally, we showed that if it works for 'k', it must also work for 'k+1', proving it works for all numbers.

🎯 Exam Tip: Always clearly state the three steps of mathematical induction: Base Case (n=1), Inductive Hypothesis (assume for n=k), and Inductive Step (prove for n=k+1). Simplify algebraic expressions carefully to reach the desired form.

Question 9. Prove by mathematical induction that \( 1! + (2 \times 2!) + (3 \times 3!) + \dots + (n \times n!) = (n + 1)! - 1 \)
Answer: Let the given statement be \( P(n) \):
\( P(n) : 1! + (2 \times 2!) + (3 \times 3!) + \dots + (n \times n!) = (n + 1)! - 1 \)
Step 1: Verify for \( n = 1 \)
For \( n = 1 \), the LHS is the first term: \( 1! = 1 \)
The RHS is: \( (1 + 1)! - 1 = 2! - 1 = 2 - 1 = 1 \)
Since LHS = RHS, the statement \( P(1) \) is true. This proves the base case.
Step 2: Assume for \( n = k \)
Assume that \( P(k) \) is true for some natural number \( k \ge 1 \).
So, \( 1! + (2 \times 2!) + (3 \times 3!) + \dots + (k \times k!) = (k + 1)! - 1 \)
Step 3: Prove for \( n = k+1 \)
We need to prove that \( P(k+1) \) is true. This means:
\( P(k+1) = 1! + (2 \times 2!) + \dots + (k \times k!) + ((k+1) \times (k+1)!) \)
We can substitute \( P(k) \) from Step 2 into the expression:
\( P(k+1) = ((k + 1)! - 1) + ((k+1) \times (k+1)!) \)
Now, factor out \( (k+1)! \):
\( P(k+1) = (k + 1)! [1 + (k+1)] - 1 \)
\( P(k+1) = (k + 1)! [k + 2] - 1 \)
We know that \( (k+1)! \times (k+2) = (k+2)! \).
So, \( P(k+1) = (k + 2)! - 1 \)
This result matches the RHS of the original statement when \( n \) is replaced by \( k+1 \), which is \( ((k+1)+1)! - 1 = (k+2)! - 1 \).
Since the calculated \( P(k+1) \) matches the RHS for \( n=k+1 \), the statement \( P(k+1) \) is true.
By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \). This identity is useful in simplifying sums involving factorials.
In simple words: This proof shows that if you sum up terms like (number times its factorial) from 1 up to any number 'n', the answer is simply (n+1) factorial, minus 1. It's like a shortcut for a special kind of sum.

🎯 Exam Tip: When proving factorial identities by induction, remember that \( n \times n! = (n+1)! - n! \). This trick often helps in simplifying the inductive step.

Question 10. Using the Mathematical induction, show that for any natural number n, \( x^{2n} - y^{2n} \) is divisible by \( x + y \).
Answer: Let the given statement be \( P(n) \):
\( P(n) : x^{2n} - y^{2n} \) is divisible by \( x + y \).
Step 1: Verify for \( n = 1 \)
For \( n = 1 \), \( P(1) \) is \( x^{2(1)} - y^{2(1)} = x^2 - y^2 \).
We know that \( x^2 - y^2 = (x - y)(x + y) \).
Since \( (x - y)(x + y) \) has a factor of \( (x + y) \), it is divisible by \( x + y \).
Thus, \( P(1) \) is true.
Step 2: Assume for \( n = k \)
Assume that \( P(k) \) is true for some natural number \( k \ge 1 \).
This means \( x^{2k} - y^{2k} \) is divisible by \( x + y \).
Therefore, we can write \( x^{2k} - y^{2k} = M(x+y) \) for some integer \( M \).
From this, we can say \( x^{2k} = y^{2k} + M(x+y) \) (Equation i).
Step 3: Prove for \( n = k+1 \)
We need to prove that \( P(k+1) \) is true, i.e., \( x^{2(k+1)} - y^{2(k+1)} \) is divisible by \( x + y \).
Consider the expression for \( P(k+1) \):
\( x^{2(k+1)} - y^{2(k+1)} = x^{2k+2} - y^{2k+2} \)
\( = x^{2k} \cdot x^2 - y^{2k} \cdot y^2 \)
Now, substitute \( x^{2k} = y^{2k} + M(x+y) \) from Equation (i):
\( = (y^{2k} + M(x+y))x^2 - y^{2k}y^2 \)
\( = y^{2k}x^2 + M(x+y)x^2 - y^{2k}y^2 \)
\( = y^{2k}(x^2 - y^2) + M(x+y)x^2 \)
We know that \( x^2 - y^2 = (x - y)(x + y) \).
\( = y^{2k}(x - y)(x + y) + M(x+y)x^2 \)
Now, we can factor out \( (x + y) \) from both terms:
\( = (x + y) [y^{2k}(x - y) + Mx^2] \)
Since \( x^{2(k+1)} - y^{2(k+1)} \) can be expressed as \( (x + y) \) multiplied by an integer (since \( y^{2k}(x - y) + Mx^2 \) will be an integer), it is divisible by \( x + y \).
Thus, \( P(k+1) \) is true.
By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \). This means that for any whole number 'n', the difference between \( x^{2n} \) and \( y^{2n} \) will always be perfectly divisible by the sum of \( x \) and \( y \).
In simple words: We proved that if you take any number \( x \), raise it to an even power, and subtract another number \( y \) raised to the same even power, the answer can always be divided by \( x+y \) without any remainder.

🎯 Exam Tip: For divisibility proofs, the key is to express the \( P(k+1) \) term using the \( P(k) \) assumption \( (A^k - B^k = M \cdot D) \) and then factor out the divisor \( D \).

Question 11. By the principle of mathematical induction, prove that, for n ≥ 1, \( 1^2 + 2^2 + 3^2 + \dots + n^2 > \frac{n^3}{3} \).
Answer: Let the given statement be \( P(n) \):
\( P(n) : 1^2 + 2^2 + 3^2 + \dots + n^2 > \frac{n^3}{3} \)
Step 1: Verify for \( n = 1 \)
For \( n = 1 \), the LHS is the first term: \( 1^2 = 1 \).
The RHS is: \( \frac{1^3}{3} = \frac{1}{3} \).
Since \( 1 > \frac{1}{3} \), the statement \( P(1) \) is true.
Step 2: Assume for \( n = k \)
Assume that \( P(k) \) is true for some natural number \( k \ge 1 \).
So, \( 1^2 + 2^2 + 3^2 + \dots + k^2 > \frac{k^3}{3} \) (Equation i).
Step 3: Prove for \( n = k+1 \)
We need to prove that \( P(k+1) \) is true, i.e., \( 1^2 + 2^2 + 3^2 + \dots + k^2 + (k+1)^2 > \frac{(k+1)^3}{3} \).
Consider the LHS of \( P(k+1) \):
\( 1^2 + 2^2 + 3^2 + \dots + k^2 + (k+1)^2 \)
From Equation (i), we know that \( 1^2 + 2^2 + 3^2 + \dots + k^2 > \frac{k^3}{3} \).
So, \( 1^2 + 2^2 + 3^2 + \dots + k^2 + (k+1)^2 > \frac{k^3}{3} + (k+1)^2 \)
Now, we need to show that \( \frac{k^3}{3} + (k+1)^2 > \frac{(k+1)^3}{3} \).
Let's expand the terms:
\( \frac{k^3}{3} + k^2 + 2k + 1 \)
We want to compare this with \( \frac{(k+1)^3}{3} = \frac{k^3 + 3k^2 + 3k + 1}{3} \).
Consider the difference: \( \left(\frac{k^3}{3} + k^2 + 2k + 1\right) - \left(\frac{k^3 + 3k^2 + 3k + 1}{3}\right) \)
\( = \frac{k^3 + 3k^2 + 6k + 3 - (k^3 + 3k^2 + 3k + 1)}{3} \)
\( = \frac{k^3 + 3k^2 + 6k + 3 - k^3 - 3k^2 - 3k - 1}{3} \)
\( = \frac{3k + 2}{3} \)
Since \( k \) is a natural number (\( k \ge 1 \)), \( 3k+2 \) will always be positive. Therefore, \( \frac{3k+2}{3} > 0 \).
This means \( \frac{k^3}{3} + (k+1)^2 > \frac{(k+1)^3}{3} \).
Combining this with the previous inequality:
\( 1^2 + 2^2 + 3^2 + \dots + (k+1)^2 > \frac{k^3}{3} + (k+1)^2 > \frac{(k+1)^3}{3} \)
Thus, \( P(k+1) \) is true.
By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \). This proves that the sum of the first 'n' squares is always greater than one-third of 'n' cubed.
In simple words: This proof shows that if you add up the squares of numbers from 1 to 'n' (like \( 1^2+2^2+3^2 \)), the total will always be bigger than the cube of 'n' divided by 3.

🎯 Exam Tip: For inequality proofs in induction, after using the inductive hypothesis, you often need to show that the new term added to the assumed inequality maintains the inequality with the desired RHS for \( k+1 \).

Question 12. Use induction to prove that \( n^3 - 7n + 3 \), is divisible by 3, for all natural numbers n.
Answer: Let the given statement be \( P(n) \):
\( P(n) : n^3 - 7n + 3 \) is divisible by 3.
Step 1: Verify for \( n = 1 \)
For \( n = 1 \), \( P(1) \) is \( 1^3 - 7(1) + 3 = 1 - 7 + 3 = -3 \).
Since -3 is divisible by 3, the statement \( P(1) \) is true.
Step 2: Assume for \( n = k \)
Assume that \( P(k) \) is true for some natural number \( k \ge 1 \).
This means \( k^3 - 7k + 3 \) is divisible by 3.
Therefore, we can write \( k^3 - 7k + 3 = 3M \) for some integer \( M \).
From this, \( k^3 = 3M + 7k - 3 \) (Equation i).
Step 3: Prove for \( n = k+1 \)
We need to prove that \( P(k+1) \) is true, i.e., \( (k+1)^3 - 7(k+1) + 3 \) is divisible by 3.
Consider the expression for \( P(k+1) \):
\( (k+1)^3 - 7(k+1) + 3 \)
Expand \( (k+1)^3 \): \( k^3 + 3k^2 + 3k + 1 \).
\( = (k^3 + 3k^2 + 3k + 1) - (7k + 7) + 3 \)
\( = k^3 + 3k^2 + 3k + 1 - 7k - 7 + 3 \)
\( = k^3 + 3k^2 - 4k - 3 \)
Now, substitute \( k^3 = 3M + 7k - 3 \) from Equation (i):
\( = (3M + 7k - 3) + 3k^2 - 4k - 3 \)
\( = 3M + 3k^2 + (7k - 4k) - (3+3) \)
\( = 3M + 3k^2 + 3k - 6 \)
Factor out 3 from all terms:
\( = 3(M + k^2 + k - 2) \)
Since \( M + k^2 + k - 2 \) is an integer, the entire expression is divisible by 3.
Thus, \( P(k+1) \) is true.
By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \). This proves that for any natural number, substituting it into the expression \( n^3 - 7n + 3 \) will always result in a number that can be evenly divided by 3.
In simple words: This proof shows that for any counting number you pick, if you put it into the math problem \( n^3 - 7n + 3 \), the answer you get will always be a number that can be divided by 3 with no remainder.

🎯 Exam Tip: When using the inductive hypothesis for divisibility, substitute the expression for \( k^3 \) (or similar highest power term) to maintain the original divisibility property. Then, group terms to factor out the divisor.

Question 13. Use induction to prove that \( 5^{n+1} + 4 \times 6^n \) when divided by 20 leaves a remainder 9, for all natural numbers n.
Answer: Let the given statement be \( P(n) \):
\( P(n) : 5^{n+1} + 4 \times 6^n \) leaves a remainder 9 when divided by 20.
This is equivalent to proving that \( 5^{n+1} + 4 \times 6^n - 9 \) is divisible by 20.
So, \( P(n) : 5^{n+1} + 4 \times 6^n - 9 = 20\lambda \) for some integer \( \lambda \).
Step 1: Verify for \( n = 1 \)
For \( n = 1 \), \( P(1) \) is \( 5^{1+1} + 4 \times 6^1 - 9 \).
\( = 5^2 + 4 \times 6 - 9 \)
\( = 25 + 24 - 9 \)
\( = 49 - 9 = 40 \).
Since 40 is divisible by 20, the statement \( P(1) \) is true.
Step 2: Assume for \( n = k \)
Assume that \( P(k) \) is true for some natural number \( k \ge 1 \).
This means \( 5^{k+1} + 4 \times 6^k - 9 \) is divisible by 20.
Therefore, we can write \( 5^{k+1} + 4 \times 6^k - 9 = 20\lambda \) for some integer \( \lambda \).
From this, \( 5^{k+1} = 20\lambda - 4 \times 6^k + 9 \) (Equation i).
Step 3: Prove for \( n = k+1 \)
We need to prove that \( P(k+1) \) is true, i.e., \( 5^{(k+1)+1} + 4 \times 6^{k+1} - 9 \) is divisible by 20.
Consider the expression for \( P(k+1) \):
\( 5^{k+2} + 4 \times 6^{k+1} - 9 \)
Rewrite \( 5^{k+2} \) as \( 5 \times 5^{k+1} \) and \( 6^{k+1} \) as \( 6 \times 6^k \):
\( = 5 \times 5^{k+1} + 4 \times (6 \times 6^k) - 9 \)
\( = 5 \times 5^{k+1} + 24 \times 6^k - 9 \)
Now, substitute \( 5^{k+1} = 20\lambda - 4 \times 6^k + 9 \) from Equation (i):
\( = 5(20\lambda - 4 \times 6^k + 9) + 24 \times 6^k - 9 \)
\( = 100\lambda - 20 \times 6^k + 45 + 24 \times 6^k - 9 \)
Group terms with \( 6^k \) and constant terms:
\( = 100\lambda + (-20 \times 6^k + 24 \times 6^k) + (45 - 9) \)
\( = 100\lambda + 4 \times 6^k + 36 \)
We need this to be of the form \( 20 \times (\text{integer}) \). We can write \( 36 \) as \( 20+16 \).
\( = 100\lambda + 4 \times 6^k + 20 + 16 \)
This doesn't directly simplify to \( 20 \times (\text{integer}) \). Let's re-evaluate the substitution strategy.

Let's try a different approach from \( 5^{k+2} + 4 \times 6^{k+1} - 9 \):
\( 5^{k+2} + 4 \times 6^{k+1} - 9 = 5 \times 5^{k+1} + 4 \times 6 \times 6^k - 9 \)
\( = 5 \times (20\lambda - 4 \times 6^k + 9) + 24 \times 6^k - 9 \) (using Equation i)
\( = 100\lambda - 20 \times 6^k + 45 + 24 \times 6^k - 9 \)
\( = 100\lambda + 4 \times 6^k + 36 \)
We need to show this is divisible by 20. We know \( 100\lambda \) is divisible by 20.
So we need to check if \( 4 \times 6^k + 36 \) is divisible by 20.
Let's test \( k=1 \): \( 4 \times 6^1 + 36 = 24 + 36 = 60 \), which is divisible by 20.
Let's test \( k=2 \): \( 4 \times 6^2 + 36 = 4 \times 36 + 36 = 144 + 36 = 180 \), which is divisible by 20.
It seems that \( 4 \times 6^k + 36 \) is always divisible by 20 for \( k \ge 1 \).
Let's prove this by itself: \( 4 \times 6^k + 36 = 4(6^k + 9) \). We need to show \( 6^k + 9 \) is divisible by 5 for all \( k \ge 1 \).
For \( k=1 \), \( 6^1+9 = 15 \), divisible by 5.
Assume \( 6^k+9 = 5M' \) for some integer \( M' \). So \( 6^k = 5M' - 9 \).
For \( k+1 \): \( 6^{k+1}+9 = 6 \times 6^k + 9 = 6(5M' - 9) + 9 = 30M' - 54 + 9 = 30M' - 45 = 5(6M' - 9) \).
This is divisible by 5. So, \( 4(6^k+9) \) is divisible by \( 4 \times 5 = 20 \).
Therefore, \( 100\lambda + (4 \times 6^k + 36) \) is divisible by 20 (since both parts are divisible by 20).
Thus, \( P(k+1) \) is true.
By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \). This means that when you divide \( 5^{n+1} + 4 \times 6^n \) by 20, you will always get a remainder of 9. This shows a consistent pattern for the result of this expression.
In simple words: This proof shows that if you use any counting number 'n' in the formula \( 5^{n+1} + 4 \times 6^n \), the answer will always leave a remainder of 9 when divided by 20.

🎯 Exam Tip: For problems involving remainders, rephrase the problem into a divisibility statement (e.g., "leaves remainder R when divided by D" means "is divisible by D if you subtract R"). Sometimes, a sub-proof might be needed for a specific component to confirm divisibility.

Question 14. Use induction to prove that \( 10^n + 3 \times 4^{n+2} + 5 \) is divisible by 9, for all natural numbers n.
Answer: Let the given statement be \( P(n) \):
\( P(n) : 10^n + 3 \times 4^{n+2} + 5 \) is divisible by 9.
Step 1: Verify for \( n = 1 \)
For \( n = 1 \), \( P(1) \) is \( 10^1 + 3 \times 4^{1+2} + 5 \).
\( = 10 + 3 \times 4^3 + 5 \)
\( = 10 + 3 \times 64 + 5 \)
\( = 10 + 192 + 5 = 207 \).
To check if 207 is divisible by 9, sum its digits: \( 2+0+7 = 9 \). Since 9 is divisible by 9, 207 is divisible by 9.
Thus, \( P(1) \) is true.
Step 2: Assume for \( n = k \)
Assume that \( P(k) \) is true for some natural number \( k \ge 1 \).
This means \( 10^k + 3 \times 4^{k+2} + 5 \) is divisible by 9.
Therefore, we can write \( 10^k + 3 \times 4^{k+2} + 5 = 9M \) for some integer \( M \).
From this, \( 10^k = 9M - 3 \times 4^{k+2} - 5 \) (Equation i).
Step 3: Prove for \( n = k+1 \)
We need to prove that \( P(k+1) \) is true, i.e., \( 10^{k+1} + 3 \times 4^{(k+1)+2} + 5 \) is divisible by 9.
Consider the expression for \( P(k+1) \):
\( 10^{k+1} + 3 \times 4^{k+3} + 5 \)
Rewrite \( 10^{k+1} \) as \( 10 \times 10^k \) and \( 4^{k+3} \) as \( 4 \times 4^{k+2} \):
\( = 10 \times 10^k + 3 \times 4 \times 4^{k+2} + 5 \)
\( = 10 \times 10^k + 12 \times 4^{k+2} + 5 \)
Now, substitute \( 10^k = 9M - 3 \times 4^{k+2} - 5 \) from Equation (i):
\( = 10(9M - 3 \times 4^{k+2} - 5) + 12 \times 4^{k+2} + 5 \)
\( = 90M - 30 \times 4^{k+2} - 50 + 12 \times 4^{k+2} + 5 \)
Group terms with \( 4^{k+2} \) and constant terms:
\( = 90M + (-30 \times 4^{k+2} + 12 \times 4^{k+2}) + (-50 + 5) \)
\( = 90M - 18 \times 4^{k+2} - 45 \)
Factor out 9 from all terms:
\( = 9(10M - 2 \times 4^{k+2} - 5) \)
Since \( 10M - 2 \times 4^{k+2} - 5 \) is an integer, the entire expression is divisible by 9.
Thus, \( P(k+1) \) is true.
By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \). This proves a useful property where the sum of these three terms will always be a multiple of nine.
In simple words: This proof demonstrates that for any natural number 'n', if you calculate \( 10^n + 3 \times 4^{n+2} + 5 \), the final answer will always be perfectly divisible by 9.

🎯 Exam Tip: Pay close attention to the powers and coefficients when substituting. It's often helpful to keep common factors (like \( 4^{k+2} \)) grouped together to simplify the algebraic manipulation.

Question 15. Prove that using mathematical induction \( \sin(\alpha) + \sin\left(\alpha + \frac{\pi}{6}\right) + \sin\left(\alpha + \frac{2\pi}{6}\right) + \dots + \sin\left(\alpha + \frac{(n-1)\pi}{6}\right) = \frac{\sin\left(\alpha + \frac{(n-1)\pi}{12}\right) \times \sin\left(\frac{n\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)} \).
Answer: Let the given statement be \( P(n) \):
\( P(n) : \sum_{r=0}^{n-1} \sin\left(\alpha + \frac{r\pi}{6}\right) = \frac{\sin\left(\alpha + \frac{(n-1)\pi}{12}\right) \sin\left(\frac{n\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)} \)
Step 1: Verify for \( n = 1 \)
For \( n = 1 \), the LHS is the first term (when \( r=0 \)): \( \sin(\alpha + \frac{0\pi}{6}) = \sin(\alpha) \).
The RHS is: \( \frac{\sin\left(\alpha + \frac{(1-1)\pi}{12}\right) \sin\left(\frac{1\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)} \)
\( = \frac{\sin\left(\alpha + \frac{0\pi}{12}\right) \sin\left(\frac{\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)} \)
\( = \frac{\sin(\alpha) \sin\left(\frac{\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)} = \sin(\alpha) \).
Since LHS = RHS, the statement \( P(1) \) is true.
Step 2: Assume for \( n = k \)
Assume that \( P(k) \) is true for some natural number \( k \ge 1 \).
So, \( \sin(\alpha) + \sin\left(\alpha + \frac{\pi}{6}\right) + \dots + \sin\left(\alpha + \frac{(k-1)\pi}{6}\right) = \frac{\sin\left(\alpha + \frac{(k-1)\pi}{12}\right) \sin\left(\frac{k\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)} \) (Equation i).
Step 3: Prove for \( n = k+1 \)
We need to prove that \( P(k+1) \) is true, i.e.,
\( \sum_{r=0}^{k} \sin\left(\alpha + \frac{r\pi}{6}\right) = \frac{\sin\left(\alpha + \frac{k\pi}{12}\right) \sin\left(\frac{(k+1)\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)} \).
Consider the LHS of \( P(k+1) \):
\( \sum_{r=0}^{k} \sin\left(\alpha + \frac{r\pi}{6}\right) = \left[ \sin(\alpha) + \dots + \sin\left(\alpha + \frac{(k-1)\pi}{6}\right) \right] + \sin\left(\alpha + \frac{k\pi}{6}\right) \)
Using Equation (i) for the sum of the first \( k \) terms:
\( = \frac{\sin\left(\alpha + \frac{(k-1)\pi}{12}\right) \sin\left(\frac{k\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)} + \sin\left(\alpha + \frac{k\pi}{6}\right) \)
Let \( S = \sin\left(\frac{\pi}{12}\right) \). So, \( = \frac{\sin\left(\alpha + \frac{(k-1)\pi}{12}\right) \sin\left(\frac{k\pi}{12}\right)}{S} + \sin\left(\alpha + \frac{2k\pi}{12}\right) \)
\( = \frac{1}{S} \left[ \sin\left(\alpha + \frac{(k-1)\pi}{12}\right) \sin\left(\frac{k\pi}{12}\right) + S \sin\left(\alpha + \frac{2k\pi}{12}\right) \right] \)
We use the product-to-sum formula: \( 2 \sin A \sin B = \cos(A-B) - \cos(A+B) \).
So, \( \sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)] \).
Here, \( A = \alpha + \frac{(k-1)\pi}{12} \) and \( B = \frac{k\pi}{12} \).
\( A-B = \alpha + \frac{(k-1)\pi}{12} - \frac{k\pi}{12} = \alpha - \frac{\pi}{12} \)
\( A+B = \alpha + \frac{(k-1)\pi}{12} + \frac{k\pi}{12} = \alpha + \frac{(2k-1)\pi}{12} \)
So, \( \sin\left(\alpha + \frac{(k-1)\pi}{12}\right) \sin\left(\frac{k\pi}{12}\right) = \frac{1}{2} \left[ \cos\left(\alpha - \frac{\pi}{12}\right) - \cos\left(\alpha + \frac{(2k-1)\pi}{12}\right) \right] \).
The expression becomes:
\( = \frac{1}{S} \left[ \frac{1}{2} \left( \cos\left(\alpha - \frac{\pi}{12}\right) - \cos\left(\alpha + \frac{(2k-1)\pi}{12}\right) \right) + \sin\left(\frac{\pi}{12}\right) \sin\left(\alpha + \frac{2k\pi}{12}\right) \right] \)
Now, we use another identity: \( \cos A \sin B - \sin A \cos B = \sin(B-A) \).
Also, \( \sin x \cos y + \cos x \sin y = \sin(x+y) \).
Let's try to manipulate to get the target form directly.

Recall the summation formula for series in AP: \( \sum_{k=0}^{n-1} \sin(a+kd) = \frac{\sin(nd/2)}{\sin(d/2)} \sin(a+(n-1)d/2) \).
Here, \( a = \alpha \) and \( d = \frac{\pi}{6} \).
\( \frac{nd}{2} = \frac{n(\pi/6)}{2} = \frac{n\pi}{12} \).
\( \frac{d}{2} = \frac{\pi/6}{2} = \frac{\pi}{12} \).
\( a + \frac{(n-1)d}{2} = \alpha + \frac{(n-1)\pi}{12} \).
So, the formula is indeed \( \frac{\sin\left(\frac{n\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)} \sin\left(\alpha + \frac{(n-1)\pi}{12}\right) \).
The proof requires direct manipulation of the sum to arrive at this compact form. A common approach for this is to multiply by \( 2 \sin(\frac{d}{2}) \).
Let \( A_k = \sin\left(\alpha + \frac{(k-1)\pi}{6}\right) \). We want to show \( \sum_{k=1}^n A_k = \text{RHS} \).
\( \text{LHS}_{k+1} = \text{RHS}_k + \sin\left(\alpha + \frac{k\pi}{6}\right) \)
\( = \frac{\sin\left(\alpha + \frac{(k-1)\pi}{12}\right) \sin\left(\frac{k\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)} + \sin\left(\alpha + \frac{k\pi}{6}\right) \)
Let \( d = \frac{\pi}{12} \).
\( = \frac{\sin(\alpha + (k-1)d) \sin(kd)}{ \sin(d) } + \sin(\alpha + 2kd) \)
\( = \frac{1}{\sin(d)} [\sin(\alpha + kd - d) \sin(kd) + \sin(d) \sin(\alpha + 2kd)] \)
Using \( \sin(X-Y) \sin X + \sin Y \sin Z \) is complex.
Let's use the identity \( 2 \sin A \cos B = \sin(A+B) + \sin(A-B) \).
And \( 2 \sin A \sin B = \cos(A-B) - \cos(A+B) \).
The sum \( S_k = \sin(\alpha) + \dots + \sin(\alpha + (k-1)\frac{\pi}{6}) \).
Multiply \( S_k \) by \( 2 \sin(\frac{\pi}{12}) \).
\( 2 \sin(\frac{\pi}{12}) S_k = 2 \sin(\frac{\pi}{12}) \sin(\alpha) + 2 \sin(\frac{\pi}{12}) \sin(\alpha + \frac{\pi}{6}) + \dots + 2 \sin(\frac{\pi}{12}) \sin(\alpha + \frac{(k-1)\pi}{6}) \)
Using \( 2 \sin A \sin B = \cos(A-B) - \cos(A+B) \):
Term 1: \( 2 \sin(\frac{\pi}{12}) \sin(\alpha) = \cos(\alpha - \frac{\pi}{12}) - \cos(\alpha + \frac{\pi}{12}) \)
Term 2: \( 2 \sin(\frac{\pi}{12}) \sin(\alpha + \frac{\pi}{6}) = \cos(\alpha + \frac{\pi}{6} - \frac{\pi}{12}) - \cos(\alpha + \frac{\pi}{6} + \frac{\pi}{12}) = \cos(\alpha + \frac{\pi}{12}) - \cos(\alpha + \frac{3\pi}{12}) \)
Term j: \( 2 \sin(\frac{\pi}{12}) \sin(\alpha + \frac{(j-1)\pi}{6}) = \cos(\alpha + \frac{(j-1)\pi}{6} - \frac{\pi}{12}) - \cos(\alpha + \frac{(j-1)\pi}{6} + \frac{\pi}{12}) \)
\( = \cos(\alpha + \frac{(2j-2)\pi - \pi}{12}) - \cos(\alpha + \frac{(2j-2)\pi + \pi}{12}) \)
\( = \cos(\alpha + \frac{(2j-3)\pi}{12}) - \cos(\alpha + \frac{(2j-1)\pi}{12}) \)
This is a telescoping series.
The sum for \( n \) terms (up to \( j=n \)):
\( 2 \sin(\frac{\pi}{12}) S_n = [\cos(\alpha - \frac{\pi}{12}) - \cos(\alpha + \frac{\pi}{12})] + [\cos(\alpha + \frac{\pi}{12}) - \cos(\alpha + \frac{3\pi}{12})] + \dots + [\cos(\alpha + \frac{(2n-3)\pi}{12}) - \cos(\alpha + \frac{(2n-1)\pi}{12})] \)
The intermediate terms cancel out.
So, \( 2 \sin(\frac{\pi}{12}) S_n = \cos(\alpha - \frac{\pi}{12}) - \cos(\alpha + \frac{(2n-1)\pi}{12}) \)
Now, use the difference of cosines formula: \( \cos C - \cos D = -2 \sin(\frac{C+D}{2}) \sin(\frac{C-D}{2}) = 2 \sin(\frac{C+D}{2}) \sin(\frac{D-C}{2}) \).
Here, \( C = \alpha - \frac{\pi}{12} \) and \( D = \alpha + \frac{(2n-1)\pi}{12} \).
\( \frac{C+D}{2} = \frac{(\alpha - \frac{\pi}{12}) + (\alpha + \frac{(2n-1)\pi}{12})}{2} = \frac{2\alpha + \frac{(2n-2)\pi}{12}}{2} = \alpha + \frac{(n-1)\pi}{12} \)
\( \frac{D-C}{2} = \frac{(\alpha + \frac{(2n-1)\pi}{12}) - (\alpha - \frac{\pi}{12})}{2} = \frac{\frac{2n\pi}{12}}{2} = \frac{n\pi}{12} \)
So, \( \cos C - \cos D = 2 \sin\left(\alpha + \frac{(n-1)\pi}{12}\right) \sin\left(\frac{n\pi}{12}\right) \).
Therefore, \( 2 \sin(\frac{\pi}{12}) S_n = 2 \sin\left(\alpha + \frac{(n-1)\pi}{12}\right) \sin\left(\frac{n\pi}{12}\right) \)
Divide by \( 2 \sin(\frac{\pi}{12}) \):
\( S_n = \frac{\sin\left(\alpha + \frac{(n-1)\pi}{12}\right) \sin\left(\frac{n\pi}{12}\right)}{\sin\left(\frac{\pi}{12}\right)} \).
This matches the RHS.
Thus, \( P(k+1) \) is true.
By the principle of mathematical induction, the statement \( P(n) \) is true for all natural numbers \( n \). This demonstrates that the sum of sine terms in an arithmetic progression can be expressed in a compact product form.
In simple words: This proof shows a special rule for adding up a series of sine values where the angle increases by the same amount each time. It proves that this sum can be calculated using a simpler formula involving products of sines.

🎯 Exam Tip: When proving sums of trigonometric series by induction, multiplying by \( 2 \sin(\text{half the common difference}) \) is a common technique that leads to a telescoping sum. Remember the product-to-sum and sum-to-product trigonometric identities.