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.

Let $\begin{aligned} P(n): \cos \alpha+\cos (\alpha+\beta)+ & \cos (\alpha+2 \beta)+\supset+\cos [\alpha+(n-1) \beta] \\ = & \frac{\cos \left[\alpha+\left(\frac{n-1}{2}\right) \beta\right] \sin \left(\frac{n \beta}{2}\right)}{\sin \frac{\beta}{2}}\end{aligned}$

Step I We observe that $P(1)$

$$\begin{aligned} P(1): \cos \alpha & =\frac{\cos \left[\alpha+\left(\frac{1-1}{2}\right)\right] \beta \sin \frac{\beta}{2}}{\sin \frac{\beta}{2}}=\frac{\cos (\alpha+0) \sin \frac{\beta}{2}}{\sin \frac{\beta}{2}} \\ \cos \alpha & =\cos \alpha \end{aligned}$$

Hence, $P(1)$ is true.

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

$$\begin{aligned} & P(k): \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\supset+\cos [\alpha+(k-1) \beta] \\ &= \frac{\cos \left[\alpha+\left(\frac{k-1}{2}\right)\right] \beta \sin \frac{k \beta}{2}}{\sin \frac{\beta}{2}} \end{aligned}$$

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

$\begin{aligned} P(k+1): \cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\supset+\cos [\alpha+(k-1) \beta] \\ +\cos [\alpha+(k+1-1) \beta]=\frac{\cos \left(\alpha+\frac{k \beta}{2}\right) \sin (k+1) \frac{\beta}{2}}{\sin \frac{\beta}{2}}\end{aligned}$

$$\begin{aligned} \mathrm{LHS} & =\cos \alpha+\cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\supset+\cos [\alpha+(k-1) \beta]+\cos (\alpha+k \beta) \\ & =\frac{\cos \left[\alpha+\left(\frac{k-1}{2}\right) \beta\right] \sin \frac{k \beta}{2}}{\sin \frac{\beta}{2}}+\cos (\alpha+k \beta) \\ & =\frac{\cos \left[\alpha+\left(\frac{k-1}{2}\right) \beta\right] \sin \frac{k \beta}{2}+\cos (\alpha+k \beta) \sin \frac{\beta}{2}}{\sin \frac{\beta}{2}} \end{aligned}$$

$$\begin{aligned} & =\frac{\sin \left(\alpha+\frac{k \beta}{2}-\frac{\beta}{2}+\frac{k \beta}{2}\right)-\sin \left(\alpha+\frac{k \beta}{2}-\frac{\beta}{2}-\frac{k \beta}{2}\right)+\sin \left(\alpha+k \beta+\frac{\beta}{2}\right)-\sin \left(\alpha+k \beta-\frac{\beta}{2}\right)}{2 \sin \frac{\beta}{2}} \\ & =\frac{\sin \left(\alpha+k \beta+\frac{\beta}{2}\right)-\sin \left(\alpha-\frac{\beta}{2}\right)}{2 \sin \frac{\beta}{2}} \\ & =\frac{2 \cos \frac{1}{2}\left(\alpha+\frac{\beta}{2}+k \beta+\alpha-\frac{\beta}{2}\right) \sin \frac{1}{2}\left(\alpha+\frac{\beta}{2}+k \beta-\alpha+\frac{\beta}{2}\right)}{2 \sin \frac{\beta}{2}} \\ & =\frac{\cos \left(\frac{2 \alpha+k \beta}{2}\right) \sin \left(\frac{k \beta+\beta}{2}\right)}{\sin \frac{\beta}{2}}=\frac{\cos \left(\alpha+\frac{k \beta}{2}\right) \sin (k+1) \frac{\beta}{2}}{\sin \frac{\beta}{2}}=\text { RHS } \end{aligned}$$

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

$$ \begin{aligned} & \text { Let } P(n): \cos \theta \cos 2 \theta \supset \cos 2^{n-1} \theta=\frac{\sin 2^n \theta}{2^n \sin \theta} \\ & \text { Step I For } n=1, P(1): \cos \theta=\frac{\sin 2^1 \theta}{2^1 \sin \theta} \\ & =\frac{\sin 2 \theta}{2 \sin \theta}=\frac{2 \sin \theta \cos \theta}{2 \sin \theta}=\cos \theta \end{aligned}$$

which is true.

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

$P(k): \cos \theta \cdot \cos 2 \theta \cdot \cos ^2 \theta \supset \cos ^{k-1} \theta=\frac{\sin 2^k \theta}{2^k \sin \theta}$ is true.

$$\begin{aligned} &\text { Step III To prove } P(k+1) \text { is true. }\\ &\begin{aligned} & P(k+1): \cos \theta \cdot \cos 2 \theta \cdot \cos 2^2 \theta \supset \cos 2^{k-1} \theta \cdot \cos 2^k \theta \\ &=\frac{\sin 2^k \theta}{2^k \sin \theta} \cdot \cos 2^k \theta \\ &=\frac{2 \sin 2^k \theta \cdot \cos 2^k \theta}{2 \cdot 2^k \sin \theta} \\ &=\frac{\sin 2 \cdot 2^k \theta}{2^{k+1} \sin \theta}=\frac{\sin 2^{(k+1)} \theta}{2^{k+1} \sin \theta} \end{aligned} \end{aligned}$$

which is true.

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