<s>[INST] Roots of a polynomial whose coefficients are ratios of binomial coefficients 
Prove that $\left\{\cot^2\left(\dfrac{k\pi}{2n+1}\right)\right\}_{k=1}^{n}$ are the roots of the equation
$$x^n-\dfrac{\dbinom{2n+1}{3}}{\dbinom{2n+1}{1}}x^{n-1} + \dfrac{\dbinom{2n+1}{5}}{\dbinom{2n+1}{1}}x^{n-2} - \ldots \ldots \ldots +  \dfrac{(-1)^{n}}{\dbinom{2n+1}{1}} =0 $$
Hence prove that
$$\sum_{r=1}^{\infty} \dfrac{1}{r^2}=\dfrac{\pi^{2}}{6}$$

I was stumped on the first sight. Then I tried using complex numbers but it was in vein. I further tried simplifying the equation, but since it contains only half of the binomial coefficients, I wasn't able to get a simpler equation.
Any help will be appreciated.
Thanks.
 [/INST] Like Sum of tangent functions where arguments are in specific arithmetic series,
$$\displaystyle\tan(2n+1)x=\dfrac{\binom{2n+1}1\tan x-\binom{2n+1}3\tan^3x+\cdots}{1-\binom{2n+1}2\tan^2x+\cdots}$$
Multiplying the denominator & the numerator by $\cot^{2n+1}x$ we get, 
$$\displaystyle\tan(2n+1)x=\dfrac{\binom{2n+1}1\cot^{2n}x-\binom{2n+1}3\cot^{2n-2}x+\cdots}{\cot^{2n+1}x-\binom{2n+1}2\cot^{2n-1}x+\cdots}$$
If $\tan(2n+1)x=0,(2n+1)x=r\pi$ where $r$ is any integer
$\implies x=\dfrac{r\pi}{2n+1}$ where $r=-n,-(n-1),\cdots,0,1,\cdots,n\pmod{2n+1}$
So, the finite roots of $$\binom{2n+1}1y^{2n}-\binom{2n+1}3y^{2(n-1)}+\cdots=0$$
are $y=\cot x,$ where $x=\dfrac{r\pi}{2n+1}$ where $r=\pm1,\pm2\cdots,\pm n\pmod{2n+1}$
But $\cot(-A)=-\cot A$
So, the finite roots of $$\binom{2n+1}1z^n-\binom{2n+1}3z^{(n-1)}+\cdots=0$$
are $z=\cot^2x,$ where $x=\dfrac{r\pi}{2n+1}$ where $r=1,2\cdots,n\pmod{2n+1}$
The rest has already been taken care of in the other answer
</s>