Using mathematical induction,the numbers $a_n$ are defined by:
$a_0 = 1, a_{n+1} = 3n^2 + n + a_n, (n \geq 0)$.
Then,$a_n$ is equal to:

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

Prove the following by using the principle of mathematical induction for all $n \in N$:
$1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \ldots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3}$

Use the Principle of Mathematical Induction to show that for a sequence $d_{1}, d_{2}, d_{3}, \ldots$ defined by $d_{1}=2$ and $d_{k}=\frac{d_{k-1}}{k}$ for all $k \geq 2$,the general term is $d_{n}=\frac{2}{n!}$ for all $n \in N$.

For all $n \ge 1$,prove that: $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots + \frac{1}{n(n+1)} = \frac{n}{n+1}$

For all $n \in \mathbb{N}$,if $1^2+2^2+3^2+\ldots+n^2 > x$,then $x=$