A sequence $b_0, b_1, b_2, \ldots$ is defined by letting $b_0=5$ and $b_k=4+b_{k-1}$, for all natural numbers $k$. Show that $b_n=5+4 n$, for all natural number $n$ using mathematical induction.
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.
A sequence $d_1, d_2, d_3, \ldots$ is defined by letting $d_1=2$ and $d_k=\frac{d_{k-1}}{k}$, for all natural numbers, $k \geq 2$. Show that $d_n=\frac{2}{n!}$, for all $n \in N$.
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.
Prove that for all $n \in N$ $$ \begin{aligned} \cos \alpha+ & \cos (\alpha+\beta)+\cos (\alpha+2 \beta)+\supset+\cos [\alpha+(n-1) \beta] \\ = & \frac{\cos \left[\alpha+\left(\frac{n-1}{2}\right) \beta\right] \sin \left(\frac{n \beta}{2}\right)}{\sin \frac{\beta}{2}} \end{aligned}$$
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.
Prove that $\cos \theta \cos 2 \theta \cos 2^2 \theta D \cos 2^{n-1} \theta=\frac{\sin 2^n \theta}{2^n \sin \theta}, \forall n \in N$
$$ \begin{aligned} & \text { Let } P(n): \cos \theta \cos 2 \theta \supset \cos 2^{n-1} \theta=\frac{\sin 2^n \theta}{2^n \sin \theta} \\ & \text { Step I For } n=1, P(1): \cos \theta=\frac{\sin 2^1 \theta}{2^1 \sin \theta} \\ & =\frac{\sin 2 \theta}{2 \sin \theta}=\frac{2 \sin \theta \cos \theta}{2 \sin \theta}=\cos \theta \end{aligned}$$
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.
$$ \begin{aligned} & \text { Prove that, } \sin \theta+\sin 2 \theta+\sin 3 \theta+\supset+\sin n \theta=\frac{\frac{\sin n \theta}{2} \sin \frac{(n+1)}{2} \theta}{\sin \frac{\theta}{2}} \text {, } \\ & \text { for all } n \in N \text {. } \end{aligned} $$
Consider the given statement
$$\begin{aligned} & P(n): \sin \theta+\sin 2 \theta+\sin 3 \theta+\supset+\sin n \theta \\ & =\frac{\sin \frac{n \theta}{2} \sin \frac{(n+1) \theta}{2}}{\sin \frac{\theta}{2}} \text {, for all } n \in N \end{aligned}$$
Step I We observe that $P(1)$ is
$$\begin{aligned} P(1): \sin \theta & =\frac{\sin \frac{\theta}{2} \cdot \sin \frac{(1+1)}{2} \theta}{\sin \frac{\theta}{2}}=\frac{\sin \frac{\theta}{2} \cdot \sin \theta}{\sin \frac{\theta}{2}} \\ \sin \theta & =\sin \theta \end{aligned}$$
Hence, $P(1)$ is true.
Step II Assume that $P(n)$ is true, for $n=k$.
$$\begin{aligned} P(k): \sin \theta+\sin 2 \theta & +\sin 3 \theta+\supset+\sin k \theta \\ & =\frac{\sin \frac{k \theta}{2} \sin \left(\frac{k+1}{2}\right) \theta}{\sin \frac{\theta}{2}} \text { is true. } \end{aligned}$$
Step III Now, to prove $P(k+1)$ is true.
$$\begin{gathered} P(k+1): \sin \theta+\sin 2 \theta+\sin 3 \theta+\supset+\sin k \theta+\sin (k+1) \theta \\ =\frac{\sin \frac{(k+1) \theta}{2} \sin \left(\frac{k+1+1}{2}\right) \theta}{\sin \frac{\theta}{2}} \end{gathered}$$
$$ \begin{aligned} \text { LHS } & =\sin \theta+\sin 2 \theta+\sin 3 \theta+\supset+\sin k \theta+\sin (k+1) \theta \\ & =\frac{\sin \frac{k \theta}{2} \sin \left(\frac{k+1}{2}\right) \theta}{\sin \frac{\theta}{2}}+\sin (k+1) \theta=\frac{\sin \frac{k \theta}{2} \sin \left(\frac{k+1}{2}\right) \theta+\sin (k+1) \theta \cdot \sin \frac{\theta}{2}}{\sin \frac{\theta}{2}} \\ & =\frac{\cos \left[\frac{k \theta}{2}-\left(\frac{k+1}{2}\right) \theta\right]-\cos \left[\frac{k \theta}{2}+\left(\frac{k+1}{2}\right) \theta\right]+\cos \left[(k+1) \theta-\frac{\theta}{2}\right]-\cos \left[(k+1) \theta+\frac{\theta}{2}\right]}{2 \sin \frac{\theta}{2}} \\ & =\frac{\cos \frac{\theta}{2}-\cos \left(k \theta+\frac{\theta}{2}\right)+\cos \left(k \theta+\frac{\theta}{2}\right)-\cos \left(k \theta+\frac{3 \theta}{2}\right)}{2 \sin \frac{\theta}{2}} \\ & =\frac{\cos \frac{\theta}{2}-\cos \left(k \theta+\frac{3 \theta}{2}\right)}{2 \sin \frac{\theta}{2}}=\frac{2 \sin \frac{1}{2}\left(\frac{\theta}{2}+k \theta+\frac{3 \theta}{2}\right) \cdot \sin \frac{1}{2}\left(k \theta+\frac{3 \theta}{2}-\frac{\theta}{2}\right)}{2 \sin \frac{\theta}{2}} \\ & =\frac{\sin \left(\frac{k \theta+2 \theta}{2}\right) \cdot \sin \left(\frac{k \theta+\theta}{2}\right)}{\sin \frac{\theta}{2}}=\frac{\sin (k+1) \frac{\theta}{2} \cdot \sin (k+1+1) \frac{\theta}{2}}{\sin \frac{\theta}{2}} \end{aligned} $$ So, $P(k+1)$ is true, whenever $P(k)$ is true. Hence, $P(n)$ is true.