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.