Mathematics Extension 2

The $n$th pentagonal number can be represented by the arithmetic series

Hence show that

for $n\in\mathbb{Z}^+$.

Hence prove by induction for all $a_1,a_2,a_3,\ldots,a_n>0$ that
$(a_1+a_2+\cdots+a_n)\left(\frac1{a_1}+\frac1{a_2}+\cdots+\frac1{a_n}\right)\ge n^2.$

(Note: this has already been proven for $n=1$ and $n=2$.)

The numbers $a_n$, for integers $n\ge1$, are defined as
$a_1=\sqrt2,\quad a_2=\sqrt{2+\sqrt2},\quad a_3=\sqrt{2+\sqrt{2+\sqrt2}},$
and so on.

These numbers satisfy the relation $a_{n+1}^2=2+a_n$, for $n\ge1$. (Do NOT prove this.)

Use mathematical induction to prove that
$a_n=2\cos\frac{\pi}{2^{n+1}},$ for all integers .

A recurrence relation is defined by
$u_{n+1}=\frac{3u_n-1}{4u_n-1},\qquad u_1=1.$
Use mathematical induction to prove that
$u_n=\frac n{2n-1}\quad\text{for }n\ge1.$

Let $J_n=\int_0^1x^ne^{-x}\,dx$, where $n$ is a non-negative integer. Using parts (i) and (iii), show by mathematical induction, or otherwise, that for all $n\ge0$,
$J_n=n!-\frac{n!}{e}\sum_{r=0}^{n}\frac1{r!}.$

Prove by mathematical induction that, for $n\geq2$,
$\frac1{2^2}+\frac1{3^2}+\cdots+\frac1{n^2}<\frac{n-1}{n}.$