For all natural numbers $n$,$3(5^{2n+1}) + 2^{3n+1}$ is divisible by:

Prove the following by using the principle of mathematical induction for all $n \in N:$
$2^{3n}-1$ is divisible by $7$.

Prove the following by using the principle of mathematical induction for all $n \in N$:
$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \dots \left(1+\frac{2n+1}{n^{2}}\right)=(n+1)^{2}$

Let $P(n): 2+2^2+2^3+\ldots+2^n=2^{n+1}-2, n \in N$. Then,

Prove the following by using the principle of mathematical induction for all $n \in N$:
$\frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \frac{1}{7 \times 9} + \ldots + \frac{1}{(2n+1)(2n+3)} = \frac{n}{3(2n+3)}$

Prove the statement by the Principle of Mathematical Induction: $n^{3}-n$ is divisible by $6$ for each natural number $n \geq 2$.