CBSE Class 11 Mathematics Principle Of Mathematical Induction Worksheet Set 05

CBSE Class 11 Mathematics Worksheet - Principle of Mathematical Induction (4). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Using PMI, show that \( 1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+n} = \frac{2n}{n+1} \)
Answer: Let \( P(n): 1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+n} = \frac{2n}{n+1} \)

(i) For \( n = 1 \):
\( P(1): 1 = \frac{2(1)}{1+1} = \frac{2}{2} = 1 \)
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): 1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+k} = \frac{2k}{k+1} \)

(iii) To prove \( P(k + 1) \) is true:
\( P(k + 1): 1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+k} + \frac{1}{1+2+3+\dots+k+(k+1)} = \frac{2(k+1)}{(k+1)+1} = \frac{2k+2}{k+2} \)

Taking L.H.S.:
\( 1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+k} + \frac{1}{1+2+3+\dots+(k+1)} \)
\( = \frac{2k}{k+1} + \frac{1}{1+2+3+\dots+(k+1)} \qquad \text{ \{from } P(k)\text{\}} \)
\( = \frac{2k}{k+1} + \frac{1}{\frac{(k+1)(k+2)}{2}} \qquad \left[\because 1+2+\dots+r = \frac{r(r+1)}{2}\right] \)
\( = \frac{2k}{k+1} + \frac{2}{(k+1)(k+2)} \)
\( = \frac{2k(k+2) + 2}{(k+1)(k+2)} \)
\( = \frac{2k^2 + 4k + 2}{(k+1)(k+2)} \)
\( = \frac{2(k^2 + 2k + 1)}{(k+1)(k+2)} \)
\( = \frac{2(k+1)^2}{(k+1)(k+2)} \)
\( = \frac{2(k+1)}{k+2} = \text{R.H.S.} \)

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in \mathbb{N} \).

Question. By PMI show that, \( \left(1 + \frac{3}{1}\right) \left(1 + \frac{5}{4}\right) \left(1 + \frac{7}{9}\right)\dots\left(1 + \frac{2n+1}{n^2}\right) = (n + 1)^2 \)
Answer: Let \( P(n): \left(1 + \frac{3}{1}\right) \left(1 + \frac{5}{4}\right) \left(1 + \frac{7}{9}\right)\dots\left(1 + \frac{2n+1}{n^2}\right) = (n + 1)^2 \)

(i) For \( n = 1 \):
\( P(1): \left(1 + \frac{3}{1}\right) = (1 + 1)^2 \Rightarrow 4 = 4 \)
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): \left(1 + \frac{3}{1}\right) \left(1 + \frac{5}{4}\right) \left(1 + \frac{7}{9}\right)\dots\left(1 + \frac{2k+1}{k^2}\right) = (k + 1)^2 \)

(iii) To prove \( P(k + 1) \) is true:
\( P(k + 1): \left(1 + \frac{3}{1}\right) \left(1 + \frac{5}{4}\right) \left(1 + \frac{7}{9}\right)\dots\left(1 + \frac{2k+1}{k^2}\right) \left(1 + \frac{2(k+1)+1}{(k+1)^2}\right) = (k + 2)^2 \)

Taking L.H.S.:
\( \left(1 + \frac{3}{1}\right) \left(1 + \frac{5}{4}\right) \left(1 + \frac{7}{9}\right)\dots\left(1 + \frac{2k+1}{k^2}\right) \left(1 + \frac{2k+3}{(k+1)^2}\right) \)
\( = (k + 1)^2 \left[1 + \frac{2k+3}{(k+1)^2}\right] \qquad \text{ \{from } P(k)\text{\}} \)
\( = (k + 1)^2 \left[ \frac{(k+1)^2 + (2k+3)}{(k+1)^2} \right] \)
\( = (k+1)^2 + (2k+3) \)
\( = k^2 + 2k + 1 + 2k + 3 \)
\( = k^2 + 4k + 4 \)
\( = (k + 2)^2 = \text{R.H.S.} \)

\( \therefore P(k + 1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in \mathbb{N} \).

Question. Show by PMI, \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^n} = 1 - \frac{1}{2^n} \)
Answer: Let \( P(n): \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^n} = 1 - \frac{1}{2^n} \)

(i) For \( n = 1 \):
\( P(1): \frac{1}{2} = 1 - \frac{1}{2} \Rightarrow \frac{1}{2} = \frac{1}{2} \)
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^k} = 1 - \frac{1}{2^k} \)

(iii) To prove \( P(k + 1) \) is true:
\( P(k + 1): \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^{k+1}} \)

Taking L.H.S.:
\( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^k} + \frac{1}{2^{k+1}} \)
\( = 1 - \frac{1}{2^k} + \frac{1}{2^{k+1}} \qquad \text{ \{from } P(k)\text{\}} \)
\( = 1 - \left[ \frac{1}{2^k} - \frac{1}{2^{k+1}} \right] \)
\( = 1 - \left[ \frac{2 - 1}{2^{k+1}} \right] \)
\( = 1 - \frac{1}{2^{k+1}} = \text{R.H.S.} \)

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in \mathbb{N} \).

Question. By PMI, show that \( (ab)^n = a^n b^n \)
Answer: Let \( P(n): (ab)^n = a^n b^n \)

(i) For \( n = 1 \):
\( P(1): (ab)^1 = ab = a^1 b^1 \)
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): (ab)^k = a^k b^k \)

(iii) To prove \( P(k + 1) \) is true:
\( P(k + 1): (ab)^{k+1} = a^{k+1} b^{k+1} \)

Taking L.H.S.:
\( (ab)^{k+1} = (ab)^k(ab) \)
\( = (a^k b^k)(ab) \qquad \text{ \{from } P(k)\text{\}} \)
\( = (a^k \cdot a)(b^k \cdot b) \)
\( = a^{k+1} b^{k+1} = \text{R.H.S.} \)

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in \mathbb{N} \).

Question. By PMI, show that \( \sin \theta + \sin(2\theta) + \dots + \sin(n\theta) = \frac{\sin\left(\frac{n+1}{2}\right)\theta \cdot \sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \)
Answer: Let \( P(n): \sin \theta + \sin(2\theta) + \dots + \sin(n\theta) = \frac{\sin\left(\frac{n+1}{2}\theta\right)\sin\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \)

(i) For \( n = 1 \):
\( P(1): \sin \theta = \frac{\sin\left(\frac{1+1}{2}\theta\right)\sin\left(\frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \Rightarrow \sin \theta = \sin \theta \)
\( \dots P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): \sin \theta + \sin(2\theta) + \dots + \sin(k\theta) = \frac{\sin\left(\frac{k+1}{2}\theta\right)\sin\left(\frac{k\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \)

(iii) To prove \( P(k + 1) \) is true:
\( P(k + 1): \sin \theta + \sin(2\theta) + \dots + \sin(k\theta) + \sin(k+1)\theta = \frac{\sin\left(\frac{k+2}{2}\theta\right)\sin\left(\frac{k+1}{2}\theta\right)}{\sin\left(\frac{\theta}{2}\right)} \)

Taking L.H.S.:
\( \sin \theta + \sin(2\theta) + \dots + \sin(k\theta) + \sin(k+1)\theta \)
\( = \frac{\sin\left(\frac{k+1}{2}\theta\right)\sin\left(\frac{k\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} + \sin(k+1)\theta \qquad \text{ \{from } P(k)\text{\}} \)
\( = \frac{\sin\left(\frac{k+1}{2}\theta\right)\sin\left(\frac{k\theta}{2}\right) + \sin(k+1)\theta \cdot \sin\left(\frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \)
\( = \frac{\sin\left(\frac{k+1}{2}\theta\right)\sin\left(\frac{k\theta}{2}\right) + 2\sin\left(\frac{k+1}{2}\theta\right)\cos\left(\frac{k+1}{2}\theta\right)\sin\left(\frac{\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \qquad [\because \sin(2x) = 2\sin x\cos x] \)
\( = \frac{\sin\left(\frac{k+1}{2}\theta\right) \left[ \sin\left(\frac{k\theta}{2}\right) + 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{k+1}{2}\theta\right) \right]}{\sin\left(\frac{\theta}{2}\right)} \)
\( = \frac{\sin\left(\frac{k+1}{2}\theta\right) \left[ \sin\left(\frac{k\theta}{2}\right) + \sin\left(\frac{k+2}{2}\theta\right) + \sin\left(-\frac{k\theta}{2}\right) \right]}{\sin\left(\frac{\theta}{2}\right)} \qquad [\because 2\sin A\cos B = \sin(A+B) + \sin(A-B)] \)
\( = \frac{\sin\left(\frac{k+1}{2}\theta\right) \left[ \sin\left(\frac{k\theta}{2}\right) + \sin\left(\frac{k+2}{2}\theta\right) - \sin\left(\frac{k\theta}{2}\right) \right]}{\sin\left(\frac{\theta}{2}\right)} \qquad [\because \sin(-x) = -\sin x] \)
\( = \frac{\sin\left(\frac{k+1}{2}\theta\right)\sin\left(\frac{k+2}{2}\theta\right)}{\sin\left(\frac{\theta}{2}\right)} = \text{R.H.S.} \)

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in \mathbb{N} \).

Question. Show by using PMI, \( \cos \alpha \cdot \cos(2\alpha) \cdot \cos(4\alpha) \dots \cos(2^{n-1}\alpha) = \frac{\sin(2^n\alpha)}{2^n \sin \alpha} \)
Answer: Let \( P(n): \cos \alpha \cdot \cos(2\alpha) \cdot \cos(4\alpha) \dots \cos(2^{n-1}\alpha) = \frac{\sin(2^n\alpha)}{2^n \sin \alpha} \)

(i) For \( n = 1 \):
\( P(1): \cos \alpha = \frac{\sin(2\alpha)}{2\sin\alpha} \Rightarrow \cos \alpha = \frac{2\sin\alpha\cos\alpha}{2\sin\alpha} \Rightarrow \cos \alpha = \cos \alpha \)
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): \cos \alpha \cdot \cos(2\alpha) \cdot \cos(4\alpha) \dots \cos(2^{k-1}\alpha) = \frac{\sin(2^k\alpha)}{2^k \sin \alpha} \)

(iii) To prove \( P(k + 1) \) is true:
\( P(k + 1): \cos \alpha \cdot \cos(2\alpha) \cdot \dots \cdot \cos(2^{k-1}\alpha) \cdot \cos(2^k\alpha) = \frac{\sin(2^{k+1}\alpha)}{2^{k+1} \sin \alpha} \)

Taking L.H.S.:
\( \cos \alpha \cdot \cos(2\alpha) \cdot \dots \cdot \cos(2^{k-1}\alpha) \cdot \cos(2^k\alpha) \)
\( = \frac{\sin(2^k\alpha)}{2^k \sin \alpha} \cdot \cos(2^k\alpha) \qquad \text{ \{from } P(k)\text{\}} \)
Multiply & divide by 2:
\( = \frac{2 \sin(2^k\alpha) \cos(2^k\alpha)}{2 \cdot 2^k \sin \alpha} \)
\( = \frac{\sin(2^{k+1}\alpha)}{2^{k+1} \sin \alpha} \qquad [\because 2\sin x\cos x = \sin(2x)] \)
\( = \frac{\sin(2^{k+1}\alpha)}{2^{k+1} \sin \alpha} = \text{R.H.S.} \)

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in \mathbb{N} \).

Question. By PMI show that \( 7 + 77 + 777 + \dots + (777\dots 7) = \frac{7}{81}[10^{n+1} - 9n - 10] \)
Answer: Let \( P(n): 7 + 77 + 777 + \dots + \underbrace{(777\dots 7)}_{n\text{-digits}} = \frac{7}{81}[10^{n+1} - 9n - 10] \)

(i) For \( n = 1 \):
\( P(1): 7 = \frac{7}{81}[10^{1+1} - 9(1) - 10] \Rightarrow 7 = \frac{7}{81}(100 - 19) \Rightarrow 7 = \frac{7}{81}(81) \Rightarrow 7 = 7 \)
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): 7 + 77 + 777 + \dots + \underbrace{(777\dots 7)}_{k\text{-digits}} = \frac{7}{81}[10^{k+1} - 9k - 10] \)

(iii) To prove \( P(k + 1) \) is true:
\( P(k + 1): 7 + 77 + 777 + \dots + \underbrace{(777\dots 7)}_{k\text{-digits}} + \underbrace{(777\dots 7)}_{(k+1)\text{-digits}} = \frac{7}{81}[10^{k+2} - 9(k + 1) - 10] \)

Taking L.H.S.:
\( 7 + 77 + 777 + \dots + \underbrace{(777\dots 7)}_{k\text{-digits}} + \underbrace{(777\dots 7)}_{(k+1)\text{-digits}} \)
\( = \frac{7}{81}[10^{k+1} - 9k - 10] + \underbrace{(777\dots 7)}_{(k+1)\text{-digits}} \qquad \text{ \{from } P(k)\text{\}} \)
\( = \frac{7}{81}[10^{k+1} - 9k - 10] + \frac{7}{9}[\underbrace{(999\dots 9)}_{(k+1)\text{-times}}] \)
\( = \frac{7}{81}[10^{k+1} - 9k - 10] + \frac{7}{9}[10^{k+1} - 1] \)
\( = \frac{7}{81}[10^{k+1} - 9k - 10] + \frac{63}{81}[10^{k+1} - 1] \)
\( = \frac{7}{81}[10^{k+1} - 9k - 10 + 9 \cdot 10^{k+1} - 9] \)
\( = \frac{7}{81}[10 \cdot 10^{k+1} - 9k - 9 - 10] \)
\( = \frac{7}{81}[10^{k+2} - 9(k + 1) - 10] = \text{R.H.S.} \)

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in \mathbb{N} \).

Question. Prove by Induction \( P(n): 1 \times 1! + 2 \times 2! + 3 \times 3! + \dots + n \times n! = (n + 1)! - 1 \)
Answer: Let \( P(n): 1 \times 1! + 2 \times 2! + 3 \times 3! + \dots + n \times n! = (n + 1)! - 1 \)

(i) For \( n = 1 \):
\( P(1): 1 \times 1! = (1 + 1)! - 1 \Rightarrow 1 = 2! - 1 \Rightarrow 1 = 2 - 1 \Rightarrow 1 = 1 \)
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): 1 \times 1! + 2 \times 2! + 3 \times 3! + \dots + k \times k! = (k + 1)! - 1 \)

(iii) To prove \( P(k + 1) \) is true:
\( P(k + 1): 1 \times 1! + 2 \times 2! + 3 \times 3! + \dots + k \times k! + (k + 1)(k + 1)! = (k + 2)! - 1 \)

Taking L.H.S.:
\( 1 \times 1! + 2 \times 2! + 3 \times 3! + \dots + k \times k! + (k + 1)(k + 1)! \)
\( = (k + 1)! - 1 + (k + 1)(k + 1)! \qquad \text{ \{from } P(k)\text{\}} \)
\( = (k + 1)![1 + (k + 1)] - 1 \)
\( = (k + 1)!(k + 2) - 1 \)
\( = (k + 2)! - 1 = \text{R.H.S.} \)

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in \mathbb{N} \).

Question. Prove by PMI \( \sin \alpha + \sin(\alpha + \beta) + \sin(\alpha + 2\beta) + \dots + \sin(\alpha + (n - 1)\beta) = \frac{\sin\left(\alpha + \frac{(n-1)}{2}\beta\right) \cdot \sin\left(\frac{n\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} \)
Answer: Let \( P(n): \sin \alpha + \sin(\alpha + \beta) + \dots + \sin(\alpha + (n - 1)\beta) = \frac{\sin \left(\alpha + \frac{(n - 1)}{2}\beta\right) \cdot \sin \left(\frac{n\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \)

(i) For \( n = 1 \):
\( P(1): \sin \alpha = \frac{\sin\left(\alpha + 0\beta\right) \cdot \sin\left(\frac{\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} \Rightarrow \sin \alpha = \sin \alpha \)
\( \therefore P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): \sin \alpha + \sin(\alpha + \beta) + \dots + \sin(\alpha + (k - 1)\beta) = \frac{\sin \left(\alpha + \frac{(k - 1)}{2}\beta\right) \cdot \sin \left(\frac{k\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \)

(iii) To prove \( P(k + 1) \) is true:
\( P(k + 1): \sin \alpha + \sin(\alpha + \beta) + \dots + \sin(\alpha + k\beta) = \frac{\sin \left(\alpha + \frac{k\beta}{2}\right) \cdot \sin \left(\frac{(k + 1)\beta}{2}\right)}{\sin \left(\frac{\beta}{2}\right)} \)

Taking L.H.S.:
\( \sin \alpha + \sin(\alpha + \beta) + \dots + \sin(\alpha + (k - 1)\beta) + \sin(\alpha + k\beta) \)
\( = \frac{\sin\left(\alpha + \frac{(k-1)}{2}\beta\right) \cdot \sin\left(\frac{k\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} + \sin(\alpha + k\beta) \qquad \text{ \{from } P(k)\text{\}} \)
\( = \frac{\sin\left(\alpha + \frac{(k-1)}{2}\beta\right) \cdot \sin\left(\frac{k\beta}{2}\right) + \sin(\alpha + k\beta) \cdot \sin\left(\frac{\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} \)
\( = \frac{2\sin\left(\alpha + \frac{(k-1)}{2}\beta\right) \cdot \sin\left(\frac{k\beta}{2}\right) + 2\sin(\alpha + k\beta) \cdot \sin\left(\frac{\beta}{2}\right)}{2\sin\left(\frac{\beta}{2}\right)} \)
Using identity \( 2\sin A\sin B = \cos(A-B) - \cos(A+B) \):
\( = \frac{\left[\cos\left(\alpha - \frac{\beta}{2}\right) - \cos\left(\alpha + \frac{2k-1}{2}\beta\right)\right] + \left[\cos\left(\alpha + \frac{2k-1}{2}\beta\right) - \cos\left(\alpha + \frac{2k+1}{2}\beta\right)\right]}{2\sin\left(\frac{\beta}{2}\right)} \)
\( = \frac{\cos\left(\alpha - \frac{\beta}{2}\right) - \cos\left(\alpha + k\beta + \frac{\beta}{2}\right)}{2\sin\left(\frac{\beta}{2}\right)} \)
Using identity \( \cos C - \cos D = 2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{D-C}{2}\right) \):
\( = \frac{2\sin\left(\frac{\alpha - \frac{\beta}{2} + \alpha + k\beta + \frac{\beta}{2}}{2}\right) \cdot \sin\left(\frac{\alpha + k\beta + \frac{\beta}{2} - \alpha + \frac{\beta}{2}}{2}\right)}{2\sin\left(\frac{\beta}{2}\right)} \)
\( = \frac{2\sin\left(\alpha + \frac{k\beta}{2}\right) \cdot \sin\left(\frac{(k+1)\beta}{2}\right)}{2\sin\left(\frac{\beta}{2}\right)} \)
\( = \frac{\sin\left(\alpha + \frac{k\beta}{2}\right) \cdot \sin\left(\frac{(k+1)\beta}{2}\right)}{\sin\left(\frac{\beta}{2}\right)} = \text{R.H.S.} \)

\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in \mathbb{N} \).

Question. Prove by Induction that \( \frac{n^5}{5} + \frac{n^3}{3} + \frac{7n}{15} \) is a natural number for all \( n \in \mathbb{N} \)
Answer: Let \( P(n): \frac{n^5}{5} + \frac{n^3}{3} + \frac{7n}{15} \) is a natural number.
(i) For \( n = 1 \):
\( P(1): \frac{1^5}{5} + \frac{1^3}{3} + \frac{7(1)}{15} = \frac{1}{5} + \frac{1}{3} + \frac{7}{15} = \frac{3 + 5 + 7}{15} = \frac{15}{15} = 1 \), which is a natural number.
\( \dots P(1) \) is true.

(ii) Let \( P(k) \) be true:
\( P(k): \frac{k^5}{5} + \frac{k^3}{3} + \frac{7k}{15} = \lambda \qquad (\lambda \in \mathbb{N}) \)

(iii) To prove \( P(k+1) \) is true:
\( P(k+1): \frac{(k+1)^5}{5} + \frac{(k+1)^3}{3} + \frac{7(k+1)}{15} \) is a natural number.

Expanding the terms:
\( (k+1)^5 = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1 \)
\( (k+1)^3 = k^3 + 3k^2 + 3k + 1 \)

Substituting these into \( P(k+1) \):
\( = \frac{k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1}{5} + \frac{k^3 + 3k^2 + 3k + 1}{3} + \frac{7k + 7}{15} \)
\( = \left( \frac{k^5}{5} + \frac{k^3}{3} + \frac{7k}{15} \right) + \frac{5k^4 + 10k^3 + 10k^2 + 5k + 1}{5} + \frac{3k^2 + 3k + 1}{3} + \frac{7}{15} \)
\( = \lambda + \left( k^4 + 2k^3 + 2k^2 + k + \frac{1}{5} \right) + \left( k^2 + k + \frac{1}{3} \right) + \frac{7}{15} \)
\( = \lambda + k^4 + 2k^3 + 3k^2 + 2k + \left( \frac{1}{5} + \frac{1}{3} + \frac{7}{15} \right) \)
\( = \lambda + k^4 + 2k^3 + 3k^2 + 2k + 1 \)

Since \( \lambda \in \mathbb{N} \) and \( k \in \mathbb{N} \), \( \lambda + k^4 + 2k^3 + 3k^2 + 2k + 1 \) is also a natural number.
\( \therefore P(k+1) \) is true.
\( \therefore \) By PMI, \( P(n) \) is true for all \( n \in \mathbb{N} \).

Question. If \( P(n): 2 \cdot 4^{2n+1} + 3^{3n+1} \) is divisible by \( \lambda \) for all \( n \in \mathbb{N} \) is true, then find the value of \( \lambda \).
Answer: We have \( P(n): 2 \cdot 4^{2n+1} + 3^{3n+1} \)

For \( n = 1 \):
\( P(1): 2 \cdot 4^3 + 3^4 = 2 \times 64 + 81 = 128 + 81 = 209 \)

For \( n = 2 \):
\( P(2): 2 \cdot 4^5 + 3^7 = 2 \times 1024 + 2187 = 2048 + 2187 = 4235 \)

Now, the highest common factor (HCF) of \( P(1) \) and \( P(2) \) is:
\( \text{HCF}(209, 4235) = 11 \qquad [\because 209 = 11 \times 19 \text{ and } 4235 = 11 \times 385] \)

\( \therefore P(n) \) is divisible by 11.
\( \therefore \lambda = 11 \).

Question. If \( P(n): 49^n + 16^n + k \) is divisible by 64 is true, then find the least negative integral value of \( k \).
Answer: We have, \( P(n): 49^n + 16^n + k \)

For \( n = 1 \):
\( P(1): 49^1 + 16^1 + k = 65 + k \)

For \( P(1) \) to be divisible by 64, \( 65 + k \) must be divisible by 64.
The least negative integral value of \( k \) that satisfies this is:
\( 65 + k = 64 \Rightarrow k = -1 \)

Since \( 65 - 1 = 64 \), which is divisible by 64.
\( \therefore \) the least negative integral value of \( k \) is \( -1 \).