Let $P(n): x^n-y^n$ is divisible by $x-y$, where $x$ and $y$ are any integers with $x \neq y$.

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

$$P(1): x^1-y^1=x-y$$

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

$$P(k): x^k-y^k \text { is divisible by }(x-y) \text {. }$$

$$\therefore \quad x^k-y^k=q(x-y)$$

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

$$\begin{aligned} P(k+1) & : x^{k+1}-y^{k+1} \\ & =x^k \cdot x-y^k \cdot y \\ & =x^k \cdot x-x^k \cdot y+x^k \cdot y-y^k \cdot y \\ & =x^k(x-y)+y\left(x^k-y^k\right) \\ & =x^k(x-y)+y q(x-y) \\ & =(x-y)\left[x^k+y q\right], \text { which is divisible by }(x-y) . \quad \text { [from step II] } \end{aligned}$$

Hence, $P(k+1)$ is true whenever $P(k)$ is true. So, by the principle of mathematical induction $P(n)$ is true for any natural number $n$.

Let $P(n): n^3-n$ is divisible by 6 , for each natural number $n \geq 2$.

Step I We observe that $P(2)$ is true. $P(2):(2)^3-2$

$\Rightarrow \quad 8-2=6$, which is divisible by 6.

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

$$\begin{aligned} & P(k): k^3-k \text { is divisible by } 6 \\ & k^3-k=6 q \end{aligned}$$

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

$$\begin{aligned} P(k & +1):(k+1)^3-(k+1) \\ & =k^3+1+3 k(k+1)-(k+1) \\ & =k^3+1+3 k^2+3 k-k-1 \\ & =k^3-k+3 k^2+3 k \\ & =6 q+3 k(k+1) \quad \text{[from step II]} \end{aligned}$$

We know that, $3 k(k+1)$ is divisible by 6 for each natural number $n=k$.

So, $P(k+1)$ is true. Hence, by the principle of mathematical induction $P(n)$ is true.

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.