Let $P(n): 4^n-1$ is divisible by 3 for each natural number $n$.

Step I Now, we observe that $P(1)$ is true.

$$P(1)=4^1-1=3$$

It is clear that 3 is divisible by 3 .

Hence, $P(1)$ is true.

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

$P(k): 4^k-1$ is divisible by 3

$$x 4^k-1=3 q$$

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

$$ \begin{aligned} &\begin{aligned} & P(k+1): 4^{k+1}-1 \\ & =4^k \cdot 4-1 \\ & =4^k \cdot 3+4^k-1 \\ & =3 \cdot 4^k+3 q \quad {\left[\because 4^k-1=3 q\right]} \\ & =3\left(4^k+q\right) \end{aligned}\\ \end{aligned}$$

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

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

Let $P(n): 2^{3 n}-1$ is divisible by 7

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

$$P(1): 2^{3 \times 1}-1=2^3-1=8-1=7$$

It is clear that $P(1)$ is true.

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

$P(k): 2^{3 k}-1$ is divisible by 7.

$$\Rightarrow \quad 2^{3 k}-1=7 q$$

$$\begin{aligned} &\text { Step III Now, to prove } P(k+1) \text { is true. }\\ &\begin{aligned} P(k+1) & : 2^{3(k+1)}-1 \\ & =2^{3 k} \cdot 2^3-1 \\ & =2^{3 k}(7+1)-1 \\ & =7 \cdot 2^{3 k}+2^{3 k}-1 \\ & =7 \cdot 2^{3 k}+7 q \quad \text{[from step II]}\\ & =7\left(2^{3 k}+q\right) \end{aligned} \end{aligned}$$

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 number $n$.

Let $P(n): n^3-7 n+3$ is divisible by 3 , for all natural number $n$.

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

$$\begin{aligned} P(1) & =(1)^3-7(1)+3 \\ & =1-7+3 \\ & =-3, \text { which is divisible by } 3 . \end{aligned}$$

Hence, $P(1)$ is true.

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

$$\therefore \quad P(k)=k^3-7 k+3=3 q$$

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

$$\begin{aligned} P(k+1) & :(k+1)^3-7(k+1)+3 \\ & =k^3+1+3 k(k+1)-7 k-7+3 \\ & =k^3-7 k+3+3 k(k+1)-6 \\ & =3 q+3[k(k+1)-2] \end{aligned}$$

Hence, $P(k+1)$ is true whenever $P(k)$ is true. [from step II]

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

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.