$$ \begin{aligned} & \text { Let } P(n): \cos \theta \cos 2 \theta \supset \cos 2^{n-1} \theta=\frac{\sin 2^n \theta}{2^n \sin \theta} \\ & \text { Step I For } n=1, P(1): \cos \theta=\frac{\sin 2^1 \theta}{2^1 \sin \theta} \\ & =\frac{\sin 2 \theta}{2 \sin \theta}=\frac{2 \sin \theta \cos \theta}{2 \sin \theta}=\cos \theta \end{aligned}$$

which is true.

Step II Assume that $P(n)$ is true, for $n=k$.

$P(k): \cos \theta \cdot \cos 2 \theta \cdot \cos ^2 \theta \supset \cos ^{k-1} \theta=\frac{\sin 2^k \theta}{2^k \sin \theta}$ is true.

$$\begin{aligned} &\text { Step III To prove } P(k+1) \text { is true. }\\ &\begin{aligned} & P(k+1): \cos \theta \cdot \cos 2 \theta \cdot \cos 2^2 \theta \supset \cos 2^{k-1} \theta \cdot \cos 2^k \theta \\ &=\frac{\sin 2^k \theta}{2^k \sin \theta} \cdot \cos 2^k \theta \\ &=\frac{2 \sin 2^k \theta \cdot \cos 2^k \theta}{2 \cdot 2^k \sin \theta} \\ &=\frac{\sin 2 \cdot 2^k \theta}{2^{k+1} \sin \theta}=\frac{\sin 2^{(k+1)} \theta}{2^{k+1} \sin \theta} \end{aligned} \end{aligned}$$

which is true.

So, $P(k+1)$ is true. Hence, $P(n)$ is true.

Consider the given statement

$$\begin{aligned} & P(n): \sin \theta+\sin 2 \theta+\sin 3 \theta+\supset+\sin n \theta \\ & =\frac{\sin \frac{n \theta}{2} \sin \frac{(n+1) \theta}{2}}{\sin \frac{\theta}{2}} \text {, for all } n \in N \end{aligned}$$

Step I We observe that $P(1)$ is

$$\begin{aligned} P(1): \sin \theta & =\frac{\sin \frac{\theta}{2} \cdot \sin \frac{(1+1)}{2} \theta}{\sin \frac{\theta}{2}}=\frac{\sin \frac{\theta}{2} \cdot \sin \theta}{\sin \frac{\theta}{2}} \\ \sin \theta & =\sin \theta \end{aligned}$$

Hence, $P(1)$ is true.

Step II Assume that $P(n)$ is true, for $n=k$.

$$\begin{aligned} P(k): \sin \theta+\sin 2 \theta & +\sin 3 \theta+\supset+\sin k \theta \\ & =\frac{\sin \frac{k \theta}{2} \sin \left(\frac{k+1}{2}\right) \theta}{\sin \frac{\theta}{2}} \text { is true. } \end{aligned}$$

Step III Now, to prove $P(k+1)$ is true.

$$\begin{gathered} P(k+1): \sin \theta+\sin 2 \theta+\sin 3 \theta+\supset+\sin k \theta+\sin (k+1) \theta \\ =\frac{\sin \frac{(k+1) \theta}{2} \sin \left(\frac{k+1+1}{2}\right) \theta}{\sin \frac{\theta}{2}} \end{gathered}$$

$$ \begin{aligned} \text { LHS } & =\sin \theta+\sin 2 \theta+\sin 3 \theta+\supset+\sin k \theta+\sin (k+1) \theta \\ & =\frac{\sin \frac{k \theta}{2} \sin \left(\frac{k+1}{2}\right) \theta}{\sin \frac{\theta}{2}}+\sin (k+1) \theta=\frac{\sin \frac{k \theta}{2} \sin \left(\frac{k+1}{2}\right) \theta+\sin (k+1) \theta \cdot \sin \frac{\theta}{2}}{\sin \frac{\theta}{2}} \\ & =\frac{\cos \left[\frac{k \theta}{2}-\left(\frac{k+1}{2}\right) \theta\right]-\cos \left[\frac{k \theta}{2}+\left(\frac{k+1}{2}\right) \theta\right]+\cos \left[(k+1) \theta-\frac{\theta}{2}\right]-\cos \left[(k+1) \theta+\frac{\theta}{2}\right]}{2 \sin \frac{\theta}{2}} \\ & =\frac{\cos \frac{\theta}{2}-\cos \left(k \theta+\frac{\theta}{2}\right)+\cos \left(k \theta+\frac{\theta}{2}\right)-\cos \left(k \theta+\frac{3 \theta}{2}\right)}{2 \sin \frac{\theta}{2}} \\ & =\frac{\cos \frac{\theta}{2}-\cos \left(k \theta+\frac{3 \theta}{2}\right)}{2 \sin \frac{\theta}{2}}=\frac{2 \sin \frac{1}{2}\left(\frac{\theta}{2}+k \theta+\frac{3 \theta}{2}\right) \cdot \sin \frac{1}{2}\left(k \theta+\frac{3 \theta}{2}-\frac{\theta}{2}\right)}{2 \sin \frac{\theta}{2}} \\ & =\frac{\sin \left(\frac{k \theta+2 \theta}{2}\right) \cdot \sin \left(\frac{k \theta+\theta}{2}\right)}{\sin \frac{\theta}{2}}=\frac{\sin (k+1) \frac{\theta}{2} \cdot \sin (k+1+1) \frac{\theta}{2}}{\sin \frac{\theta}{2}} \end{aligned} $$ So, $P(k+1)$ is true, whenever $P(k)$ is true. Hence, $P(n)$ is true.

Consider the given statement

$P(n): \frac{n^5}{5}+\frac{n^3}{3}+\frac{7 n}{15}$ is a natural number, for all $n \in N$.

Step I We observe that $P(1)$ is true.

$P(1): \frac{(1)^5}{5}+\frac{1^3}{3}+\frac{7(1)}{15}=\frac{3+5+7}{15}=\frac{15}{15}=1$, which is a natural number. Hence, $P(1)$ is true.

Step II Assume that $P(n)$ is true, for $n=k$.

$P(k): \frac{k^5}{5}+\frac{k^3}{3}+\frac{7 k}{15}$ is natural number.

Step III Now, to prove $P(k+1)$ is true.

$$\begin{aligned} & \frac{(k+1)^5}{5}+\frac{(k+1)^3}{3}+\frac{7(k+1)}{15} \\ & =\frac{k^5+5 k^4+10 k^3+10 k^2+5 k+1}{5}+\frac{k^3+1+3 k(k+1)}{3}+\frac{7 k+7}{15} \\ & =\frac{k^5+5 k^4+10 k^3+10 k^2+5 k+1}{5}+\frac{k^3+1+3 k^2+3 k}{3}+\frac{7 k+7}{15} \\ & =\frac{k^5}{5}+\frac{k^3}{3}+\frac{7 k}{15}+\frac{5 k^4+10 k^3+10 k^2+5 k+1}{5}+\frac{3 k^2+3 k+1}{3}+\frac{7 k+7}{15} \\ & =\frac{k^5}{5}+\frac{k^3}{3}+\frac{7 k}{15}+k^4+2 k^3+2 k^2+k+k^2+k+\frac{1}{5}+\frac{1}{3}+\frac{7}{15} \\ & =\frac{k^5}{5}+\frac{k^3}{3}+\frac{7 k}{15}+k^4+2 k^3+3 k^2+2 k+1, \text { which is a natural number } \end{aligned}$$

So, $P(k+1)$ is true, whenever $P(k)$ is true. Hence, $P(n)$ is true.

Consider the given statement

$P(n): \frac{1}{n+1}+\frac{1}{n+2}+\supset+\frac{1}{2 n}>\frac{13}{24}$, for all natural numbers $n>1$

Step I We observe that, $P(2)$ is true,

$$\begin{aligned} P(2): \frac{1}{2+1}+\frac{1}{2+2} & >\frac{13}{24} \\ \frac{1}{3}+\frac{1}{4} & >\frac{13}{24} \\ \frac{4+3}{12} & >\frac{13}{24} \\ \frac{7}{12} & >\frac{13}{24}, \text { which is true. } \end{aligned}$$

Hence, $P(2)$ is true.

Step II Now, we assume that $P(n)$ is true, For $n=k$,

$$P(k): \frac{1}{k+1}+\frac{1}{k+2}+\supset+\frac{1}{2 k}>\frac{13}{24}$$

Step III Now, to prove $P(k+1)$ is true, we have to show that

$$ \begin{aligned} &\begin{aligned} & P(k+1): \frac{1}{k+1}+\frac{1}{k+2}+\supset+\frac{1}{2 k}+\frac{1}{2(k+1)}>\frac{13}{24} \\ & \text { Given, } \quad \frac{1}{k+1}+\frac{1}{k+2}+\supset+\frac{1}{2 k}>\frac{13}{24} \\ & \frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{2 k}+\frac{1}{2(k+1)}>\frac{13}{24}+\frac{1}{2(k+1)} \\ & \frac{13}{24}+\frac{1}{2(k+1)}>\frac{13}{24} \\ & \because \quad \frac{1}{k+1}+\frac{1}{k+2}+\supset+\frac{1}{2 k}+\frac{1}{2(k+1)}>\frac{13}{24} \end{aligned}\\ \end{aligned}$$

So, $P(k+1)$ is true, whenever $P(k)$ is true. Hence, $P(n)$ is true.

Let $P(n)$ : Number of subset of a set containing $n$ distinct elements is $2^n$, for all $n \in N$.

Step I We observe that $P(1)$ is true, for $n=1$.

Number of subsets of a set contain 1 element is $2^1=2$, which is true.

Step II Assume that $P(n)$ is true for $n=k$.

$P(k)$ : Number of subsets of a set containing $k$ distinct elements is $2^k$, which is true.

Step III To prove $P(k+1)$ is true, we have to show that

$P(k+1)$ : Number of subsets of a set containing $(k+1)$ distinct elements is $2^{k+1}$.

We know that, with the addition of one element in the set, the number of subsets become double.

$\therefore$ Number of subsets of a set containing $(k+1)$ distinct elements $=2 \times 2^k=2^{k+1}$. So, $P(k+1)$ is true. Hence, $P(n)$ is true.