If $P(n): 2^{n} < n!$,then the smallest positive integer for which $P(n)$ is true,is

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 $n \in N$:
$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \dots \left(1+\frac{2n+1}{n^{2}}\right)=(n+1)^{2}$

Prove the statement by the Principle of Mathematical Induction: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}}$ for all natural numbers $n \geq 2$.

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 $n \in N$?

Prove the statement by the Principle of Mathematical Induction :
$1+5+9+\ldots+(4 n-3)=n(2 n-1)$ for all natural numbers $n$.