Let $P(n): n\left(n^2+5\right)$ is divisible by 6 , for each natural number $n$.

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

$P(1): 1\left(1^2+5\right)=6$, which is divisible by 6.

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

$$\begin{gathered} & P(k): k\left(k^2+5\right) \text { is divisible by } 6 . \\ & \therefore \quad k\left(k^2+5\right)=6 q \end{gathered}$$

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

$$\begin{aligned} P(k+1): & (k+1)\left[(k+1)^2+5\right] \\ & =(k+1)\left[k^2+2 k+1+5\right] \\ & =(k+1)\left[k^2+2 k+6\right] \\ & =k^3+2 k^2+6 k+k^2+2 k+6 \\ & =k^3+3 k^2+8 k+6 \\ & =k^3+5 k+3 k^2+3 k+6 \\ & =k\left(k^2+5\right)+3\left(k^2+k+2\right) \\ & =(6 q)+3\left(k^2+k+2\right) \end{aligned}$$

We know that, $k^2+k+2$ is divisible by 2, where, $k$ is even or odd.

Since, $P(k+1): 6 q+3\left(k^2+k+2\right)$ is divisible by 6 . So, $P(k+1)$ is true whenever $P(k)$ is true.

Hence, by the principle of mathematical induction $P(n)$ is true.

Consider the given statement

$P(n): n^2<2^n$ for all natural numbers $n \geq 5$.

Step I We observe that $P(5)$ is true

$$\begin{aligned} P(5): & 5^2<2^5 \\ & =25<32 \end{aligned}$$

Hence, $P(5)$ is true.

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

$$P(k)=k^2<2^k \text { is true. }$$

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

$$\begin{aligned} P(k+1) & :(k+1)^2<2^{k+1} \\ \text{Now,}\quad k^2<2^k & =k^2+2 k+1<2^k+2 k+1 \\ & =(k+1)^2<2^k+2 k+1 \quad \text{.... (i)}\\ \text{Now, } (2k+1)< 2^k & =2^k+2 k+1<2^k+2^k \\ & =2^k+2 k+1<2 \cdot 2^k \\ & =2^k+2 k+1<2^{k+1}\quad \text{... (ii)} \end{aligned}$$

From Eqs. (i) and (ii), we get $(k+1)^2<2^{k+1}$

So, $P(k+1)$ is true, whenever $P(k)$ is true. Hence, by the principle of mathematical induction $P(n)$ is true for all natural numbers $n \geq 5$.

Consider the statement

$P(n): 2 n<(n+2)$ ! for all natural number $n$.

Step I We observe that, $P(1)$ is true. $P(1): 2(1)<(1+2)$ !

$\Rightarrow 2<3!\Rightarrow 2<3 \times 2 \times 1 \Rightarrow 2<6$

Hence, $P(1)$ is true.

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

$$P(k): 2 k<(k+2)!\text { is true. }$$

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

$$P(k+1): 2(k+1)<(k+1+2)!

$$\begin{aligned} \text{Now,}\quad 2 k & <(k+2)! \\ 2 k+2 & <(k+2)!+2 \\ 2(k+1) & <(k+2)!+2\quad \text{.... (i)} \end{aligned}$$

Also, $$(k+2)!+2<(k+3)!\quad \text{.... (ii)}$$

From Eqs. (i) and (ii),

$$2(k+1)<(k+1+2)!$$

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

Hence, by principle of mathematical induction $P(n)$ is true.

Consider the statement

$P(n): \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\supset+\frac{1}{\sqrt{n}}$, for all natural numbers $n \geq 2$.

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

$P(2): \sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}$, which is true.

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

$$P(k): \sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\supset+\frac{1}{\sqrt{k}} \text { is true. }$$

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

$P(k+1): \sqrt{k+1}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\supset+\frac{1}{\sqrt{k+1}}$ is true.

$$\begin{aligned} & \text { Given that, } \quad \sqrt{k}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\supset+\frac{1}{\sqrt{k}} \\ & \Rightarrow \quad \sqrt{k}+\frac{1}{\sqrt{k+1}}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\supset+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} \\ & \Rightarrow \quad \frac{(\sqrt{k})(\sqrt{k+1})+1}{\sqrt{k+1}}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}} \quad \text{... (i)}\\ & \text { If } \sqrt{k+1}<\frac{\sqrt{k} \sqrt{k+1}+1}{\sqrt{k+1}} \\ & \Rightarrow \quad k+1<\sqrt{k} \sqrt{k+1}+1 \\ & \Rightarrow \quad k<\sqrt{k(k+1)} \Rightarrow \sqrt{k}<\sqrt{k}+1 \quad \text{... (ii)} \end{aligned}$$

From Eqs. (i) and (ii),

$$\sqrt{k+1}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\supset+\frac{1}{\sqrt{k+1}}$$

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

Let $P(n): 2+4+6+\supset+2 n=n^2+n$

For all natural numbers $n$.

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

$$P(1): 2=1^2+1$$

$2=2$, which is true.

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

$$\therefore \quad P(k): 2+4+6+\supset+2 k=k^2+k$$

Step III To prove that $P(k+1)$ is true.

$$\begin{aligned} P(k+1): & 2+4+6+8+\ldots+2 k+2(k+1) \\ & =k^2+k+2(k+1) \\ & =k^2+k+2 k+2 \\ & =k^2+2 k+1+k+1 \\ & =(k+1)^2+k+1 \end{aligned}$$

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

Hence, $P(n)$ is true.