NCERT Exemplar · Class 11 Mathematics Principle of Mathematical Induction

Let \(\displaystyle P(n)\): \(\displaystyle \cos\theta\cos2\theta\cos2^2\theta\cdots\cos2^{n-1}\theta = \dfrac{\sin 2^n\theta}{2^n\sin\theta}\). Base case: \[\frac{\sin 2\theta}{2\sin\theta} = \frac{2\sin\theta\cos\theta}{2\sin\theta} = \cos\theta \] Step: assume \(\displaystyle P(k)\) and multiply both sides by \(\displaystyle \cos 2^k\theta\). \[\cos\theta\cos2\theta\cdots\cos2^{k-1}\theta\cdot\cos2^k\theta = \frac{\sin 2^k\theta\,\cos 2^k\theta}{2^k\sin\theta} \quad \text{(hypothesis)} \] \[= \frac{2\sin 2^k\theta\,\cos 2^k\theta}{2^{k+1}\sin\theta} \] \[= \frac{\sin 2^{k+1}\theta}{2^{k+1}\sin\theta} \quad (\sin 2x = 2\sin x\cos x) \] Answer: \(\displaystyle P(1)\) is true and \(\displaystyle P(k)\Rightarrow P(k+1)\), so the result holds for all \(\displaystyle n\in\mathbf N\).

The printed first factor \(\displaystyle \frac{\sin n\theta}{2}\) is a misprint for \(\displaystyle \sin\frac{n\theta}{2}\); as printed, \(\displaystyle n=1\), \(\displaystyle \theta=\frac{\pi}{2}\) already fails: \[\frac{\sin^2\theta}{2\sin\dfrac{\theta}{2}} = \frac{1}{\sqrt2} \ne 1 = \sin\theta \] Let \(\displaystyle T_n = \sin\theta+\sin2\theta+\cdots+\sin n\theta\) and multiply the identity by \(\displaystyle \sin\frac{\theta}{2}\ (\ne 0)\). Let \(\displaystyle P(n)\): \[T_n\sin\tfrac{\theta}{2} = \sin\tfrac{n\theta}{2}\,\sin\tfrac{(n+1)\theta}{2} \] Base case: \[T_1\sin\tfrac{\theta}{2} = \sin\theta\,\sin\tfrac{\theta}{2} = \sin\tfrac{\theta}{2}\,\sin\tfrac{2\theta}{2} \] Step: assume \(\displaystyle P(k)\). Since \(\displaystyle T_{k+1} = T_k + \sin(k+1)\theta\): \[T_{k+1}\sin\tfrac{\theta}{2} = \sin\tfrac{k\theta}{2}\sin\tfrac{(k+1)\theta}{2} + \sin(k+1)\theta\,\sin\tfrac{\theta}{2} \quad \text{(hypothesis)} \] Use \(\displaystyle 2\sin A\sin B = \cos(A-B)-\cos(A+B)\) on each term: \[= \tfrac12\Big[\cos\tfrac{\theta}{2}-\cos\tfrac{(2k+1)\theta}{2}\Big] + \tfrac12\Big[\cos\tfrac{(2k+1)\theta}{2}-\cos\tfrac{(2k+3)\theta}{2}\Big] \] \[= \tfrac12\Big[\cos\tfrac{\theta}{2}-\cos\tfrac{(2k+3)\theta}{2}\Big] \] \[= \sin\tfrac{(k+1)\theta}{2}\,\sin\tfrac{(k+2)\theta}{2} \quad \text{(same identity, } A=\tfrac{(k+2)\theta}{2},\ B=\tfrac{(k+1)\theta}{2}\text{)} \] Answer: \(\displaystyle P(1)\) is true and \(\displaystyle P(k)\Rightarrow P(k+1)\); dividing by \(\displaystyle \sin\frac{\theta}{2}\) gives \(\displaystyle T_n = \dfrac{\sin\frac{n\theta}{2}\sin\frac{(n+1)\theta}{2}}{\sin\frac{\theta}{2}}\) for all \(\displaystyle n\in\mathbf N\).

Let \(\displaystyle f(n) = \dfrac{n^5}{5}+\dfrac{n^3}{3}+\dfrac{7n}{15}\) and \(\displaystyle P(n)\): \(\displaystyle f(n)\in\mathbf N\). Base case: \[f(1) = \frac{3}{15}+\frac{5}{15}+\frac{7}{15} = 1 \] Step: assume \(\displaystyle f(k)\in\mathbf N\). \[f(k+1)-f(k) = \frac{(k+1)^5-k^5}{5}+\frac{(k+1)^3-k^3}{3}+\frac{7}{15} \] \[= \frac{5k^4+10k^3+10k^2+5k+1}{5}+\frac{3k^2+3k+1}{3}+\frac{7}{15} \] \[= k^4+2k^3+3k^2+2k+\Big(\tfrac15+\tfrac13+\tfrac{7}{15}\Big) \] \[= k^4+2k^3+3k^2+2k+1 = (k^2+k+1)^2 \] \[f(k+1) = f(k)+(k^2+k+1)^2 \in\mathbf N \quad \text{(hypothesis)} \] Answer: \(\displaystyle P(1)\) is true and \(\displaystyle P(k)\Rightarrow P(k+1)\), so \(\displaystyle f(n)\) is a natural number for all \(\displaystyle n\in\mathbf N\).

Let \(\displaystyle S_n = \dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}\) and \(\displaystyle P(n)\): \(\displaystyle S_n>\dfrac{13}{24}\) for \(\displaystyle n\ge 2\). Base case: \[S_2 = \frac13+\frac14 = \frac{7}{12} = \frac{14}{24} > \frac{13}{24} \] Step: assume \(\displaystyle S_k>\dfrac{13}{24}\) for some \(\displaystyle k\ge 2\). \[S_{k+1} = S_k + \frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1} \] \[= S_k + \frac{1}{2k+1}-\frac{1}{2k+2} \] \[= S_k + \frac{1}{(2k+1)(2k+2)} \] \[> S_k > \frac{13}{24} \quad \text{(hypothesis)} \] Answer: \(\displaystyle P(2)\) is true and \(\displaystyle P(k)\Rightarrow P(k+1)\), so \(\displaystyle S_n>\dfrac{13}{24}\) for all natural \(\displaystyle n>1\).

Let \(\displaystyle P(n)\): a set of \(\displaystyle n\) elements has \(\displaystyle 2^n\) subsets.Base case: \[S=\{a\} \Rightarrow \text{subsets } \varnothing,\ \{a\} \] \[2 = 2^1 \]Assume \(\displaystyle P(k)\). Take \(\displaystyle S\) with \(\displaystyle k+1\) elements: \[S = A \cup \{x\}, \qquad |A| = k \] Subsets of \(\displaystyle S\) without \(\displaystyle x\) are the subsets of \(\displaystyle A\): \[2^k \quad \text{(by } P(k)\text{)} \] Subsets of \(\displaystyle S\) with \(\displaystyle x\) are \(\displaystyle B \cup \{x\}\), \(\displaystyle B \subseteq A\): \[2^k \quad \text{(by } P(k)\text{)} \] Total: \[2^k + 2^k = 2\cdot 2^k = 2^{k+1} \] So \(\displaystyle P(k) \Rightarrow P(k+1)\); \(\displaystyle P(1)\) is true.Answer: number of subsets \(\displaystyle = 2^n\) for all \(\displaystyle n \in \mathbf{N}\).

(A) \(\displaystyle 5\)Split off multiples of $\displaystyle 9$: \[10^n + 3\cdot 4^{n+2} + k = (10^n - 1) + 48\,(4^n - 1) + (49 + k) \] \[9 \mid 10^n - 1, \qquad 3 \mid 4^n - 1 \Rightarrow 9 \mid 48\,(4^n - 1) \] So only \(\displaystyle 49 + k\) matters: \[9 \mid 49 + k \Rightarrow k = 5,\ 14,\ \dots \] Check \(\displaystyle n=1\): \[10 + 192 + 5 = 207 = 9 \cdot 23 \]

(B) \(\displaystyle 17\)\[3\cdot 5^{2n+1} + 2^{3n+1} = 15\cdot 25^n + 2\cdot 8^n = 15\,(25^n - 8^n) + 17\cdot 8^n \] \[25 - 8 = 17 \mid 25^n - 8^n \] Check: \[n=1:\ 391 = 17 \cdot 23 \] \[n=2:\ 9503 = 17 \cdot 13 \cdot 43 \] $\displaystyle 23$ fails at \(\displaystyle n=2\); $\displaystyle 19$ and $\displaystyle 25$ fail at \(\displaystyle n=1\).

(A) \(\displaystyle 1\)Factor theorem: \(\displaystyle (x-k) \mid x^n - 1\) iff \[k^n - 1 = 0 \] \[k \in \mathbf{Z}^+,\ k^n = 1 \Rightarrow k = 1 \] and \(\displaystyle k=1\) works: \[x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \cdots + x + 1) \]

$\displaystyle 4$Test small \(\displaystyle n\): \[n=1:\ 2 < 1 \ \text{false}, \qquad n=2:\ 4 < 2 \ \text{false}, \qquad n=3:\ 6 < 6 \ \text{false} \] \[n=4:\ 8 < 24 \ \text{true} \] Step for \(\displaystyle k \ge 4\): assume \(\displaystyle 2k < k!\). \[(k+1)! = (k+1)\,k! > (k+1)\cdot 2k \ge 2(k+1) \] So \(\displaystyle P(4)\) holds and \(\displaystyle P(k) \Rightarrow P(k+1)\) for \(\displaystyle k \ge 4\).

False. Without a base case the implication proves nothing.Counterexample: \(\displaystyle P(n)\): \(\displaystyle n(n+1)\) is odd. \[(k+1)(k+2) = k(k+1) + 2(k+1) \] Odd plus even is odd, so \(\displaystyle P(k) \Rightarrow P(k+1)\) for every \(\displaystyle k\), yet \[P(1):\ 1\cdot 2 = 2 \ \text{is not odd} \]