Let the statement $P(n): 3 n< n!$

For $n=1,3 \times 1<1$ ! [false]

For $n=2,3 \times 2<2!\Rightarrow 6<2$ [false]

For $n=3,3 \times 3<3!\quad \Rightarrow 9<6$ [false]

For $n=4,3 \times 4<4!\quad \Rightarrow \quad 12<24$ [true]

For $n=5,3 \times 5<5!\Rightarrow 15<5 \times 4 \times 3 \times 2 \times 1 \Rightarrow 15<120$ [true]

Consider the statement

$P(n): 1^2+2^2+3^2+\supset+n^2=\frac{n(n+1)(2 n+1)}{6}$

$$\begin{array}{ll} \text { For } n=1, & 1=\frac{1(1+1)(2 \times 1+1)}{6} \\ \Rightarrow & 1=\frac{2(3)}{6} \\ \Rightarrow & 1=1 \end{array}$$

$$ \begin{array}{lr} \text { For } n=2, & 1+2^2=\frac{2(2+1)(4+1)}{6} \\ \Rightarrow & 5=\frac{30}{6} \Rightarrow 5=5 \end{array}$$

$$ \begin{aligned} &\begin{aligned} & \text { For } n=3, \quad 1+2^2+3^2=\frac{3(3+1)(7)}{6} \\ & \Rightarrow \quad 1+4+9=\frac{3 \times 4 \times 7}{6} \\ & \Rightarrow \quad 14=14 \end{aligned}\\ \end{aligned}$$

Hence, the given statement is true for all n.

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 $n$.