Let $P(n) = 3^{2n+1} + 2^{n+2}$ where $n \in N$. Then

If $P(n) = 2 + 4 + 6 + \dots + 2n$,$n \in N$,then $P(k) = k(k + 1) + 2 \implies P(k + 1) = (k + 1)(k + 2) + 2$ for all $k \in N$. So we can conclude that $P(k) = k(k + 1) + 2$ for all $k \in N$ is true. What can we conclude about $P(n) = n(n + 1) + 2$ for all $n \in N$?

Use the Principle of Mathematical Induction to prove that $\cos \theta \cos 2 \theta \cos 2^{2} \theta \ldots \cos 2^{n-1} \theta = \frac{\sin 2^{n} \theta}{2^{n} \sin \theta}$ for all $n \in N$.

Prove that $1^{2} + 2^{2} + \ldots + n^{2} > \frac{n^{3}}{3}$ for all $n \in N$.

Prove the following by using the principle of mathematical induction for all $n \in N$:
$1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \ldots + n \cdot 2^{n} = (n-1) 2^{n+1} + 2$

When $P$ is a natural number,then ${P^{n + 1}} + {(P + 1)^{2n - 1}}$ is divisible by