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.

A sequence $a_1, a_2, a_3, \ldots$ is defined by letting $a_1=3$ and $a_k=7 a_{k-1}$, for all natural numbers $k \geq 2$.

Let $\quad P(n): a_n=3 \cdot 7^{n-1}$ for all natural numbers.

Step I We observe $P(2)$ is true.

For $n=2$,

$a_2=3 \cdot 7^{2-1}=3 \cdot 7^1=21$ is true.

As $$a_1=3, a_k=7 a_{k-1}$$

$$\begin{array}{ll} \Rightarrow & a_2=7 \cdot a_{2-1}=7 \cdot a_1 \\ \Rightarrow & a_2=7 \times 3=21 \quad [\because a_1=3] \end{array}$$

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

$$P(k): a_k=3 \cdot 7^{k-1}$$

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

$$\begin{aligned} P(k+1): a_{k+1} & =3 \cdot 7^{k+1-1} \\ a_{k+1} & =7 \cdot a_{k+1-1}=7 \cdot a_k \\ & =7 \cdot 3 \cdot 7^{k-1}=3 \cdot 7^{k-1+1} \end{aligned}$$

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

Consider the given statement,

$P(n): b_n=5+4 n$, for all natural numbers given that $b_0=5$ and $b_k=4+b_{k-1}$ Step I $P(1)$ is true.

$$P(1): b_1=5+4 \times 1=9$$

As $$b_0=5, b_1=4+b_0=4+5=9$$

Hence, $P(1)$ is true.

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

$$P(k): b_k=5+4 k$$

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

$$\begin{aligned} \therefore \quad P(k+1): b_{k+1} & =5+4(k+1) \\ b_{k+1} & =4+b_{k+1-1} \\ & =4+b_k \\ & =4+5+4 k=5+4(k+1) \end{aligned}$$

So, by the mathematical induction $P(k+1)$ is true whenever $P(k)$ is true, hence $P(n)$ is true.

Let $P(n): d_n=\frac{2}{n!}, \forall n \in N$, to prove $P(2)$ is true.

Step I $$P(2): d_2=\frac{2}{2!}=\frac{2}{2 \times 1}=1$$

As, given

$$\begin{aligned} d_1 & =2 \\ d_k & =\frac{d_{k-1}}{k} \\ d_2 & =\frac{d_1}{2}=\frac{2}{2}=1 \end{aligned}$$

Hence, $P(2)$ is true.

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

$$P(k): d_k=\frac{2}{k!}$$

Step III Now, to prove that $P(k+1)$ is true, we have to show that $P(k+1): d_{k+1}=\frac{2}{(k+1)!}$

$$\begin{aligned} d_{k+1} & =\frac{d_{k+1-1}}{k}=\frac{d_k}{k} \\ & =\frac{2}{k!k}=\frac{2}{(k+1)!} \end{aligned}$$

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