Let $P(n): 3^{2 n}-1$ is divisible by 8 , for all natural numbers.

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

$$\begin{aligned} P(1): 3^{2(1)}-1 & =3^2-1 \\ & =9-1=8 \text {, which is divisible by } 8 . \end{aligned}$$

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

$$P(k): 3^{2 k}-1=8 q$$

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

$$\begin{aligned} P(k+1) & : 3^{2(k+1)}-1 \\ & =3^{2 k} \cdot 3^2-1 \\ & =3^{2 k} \cdot(8+1)-1 \\ & =8 \cdot 3^{2 k}+3^{2 k}-1 \\ & =8 \cdot 3^{2 k}+8 q \\ & =8\left(3^{2 k}+9\right)\quad \text{[from step II]} \end{aligned}$$

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

So, by the principle of mathematical induction $P(n)$ is true for all natural numbers n.

Consider the given statement is $P(n): 7^n-2^n$ is divisible by 5 , for any natural number $n$.

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

$$P(1)=7^1-2^1=5 \text {, which is divisible by } 5 \text {. }$$

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

$$P(k)=7^k-2^k=5 q$$

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

$$\begin{gathered} P(k+1): 7^{k+1}-2^{k+1} \\ =7^k \cdot 7-2^k \cdot 2\\ \begin{aligned} &\begin{aligned} & =7^k \cdot(5+2)-2^k \cdot 2 \\ & =7^k \cdot 5+2 \cdot 7^k-2^k \cdot 2 \\ & =5 \cdot 7^k+2\left(7^k-2^k\right) \\ & =5 \cdot 7^k+2(5 q) \\ & =5\left(7^k+2 q\right), \text { which is divisible by } 5 \end{aligned}\\ &\text { [from step II] } \end{aligned} \end{gathered}$$

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

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

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.