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.

Consider the given statement

$P(n): 1+2+2^2+\supset+2^n=2^{n+1}-1$, for all natural numbers $n$

Step I We observe that $P(0)$ is true.

$$\begin{aligned} P(1): 1 & =2^{0+1}-1 \\ 1 & =2^1-1 \\ 1 & =2-1 \\ 1 & =1, \text { which is true. } \end{aligned}$$

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

So, $P(k): 1+2+2^2+\supset+2^k=2^{k+1}-1$ is true.

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

$$\begin{aligned} P(k+1): & 1+2+2^2+\supset+2^k+2^{k+1} \\ & =2^{k+1}-1+2^{k+1} \\ & =2 \cdot 2^{k+1}-1 \\ & =2^{k+2}-1 \\ & =2^{(k+1)+1}-1 \end{aligned}$$

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

Hence, $P(n)$ is true.

Let $P(n): 1+5+9+\supset+(4 n-3)=n(2 n-1)$, for all natural numbers $n$.

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

$P(1): 1=1(2 \times 1-1), 1=2-1$ and $1=1$, which is true.

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

So, $P(k): 1+5+9+\supset+(4 k-3)=k(2 k-1)$ is true.

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

$$\begin{aligned} P(k+1) & : 1+5+9+\supset+(4 k-3)+4(k+1)-3 \\ & =k(2 k-1)+4(k+1)-3 \\ & =2 k^2-k+4 k+4-3 \\ & =2 k^2+3 k+1 \\ & =2 k^2+2 k+k+1 \\ & =2 k(k+1)+1(k+1) \\ & =(k+1)(2 k+1) \\ & =(k+1)[2 k+1+1-1] \\ & =(k+1)[2(k+1)-1] \end{aligned}$$

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