2016/112) Find the value of $ \displaystyle \sum_{k=1}^{n} \cot^2(\frac{\pi k}{2n+1})$

We have
$(\cos\, ma + i \sin\, ma) = (\cos\, a + i \sin\, a)^m$
= $\sum\limits_{k=0}^m {m \choose k } (i\sin\,a)^k (\cos\,a) ^{m-k} $
now compare imaginary parts on both sides to get
$\sin\, ma = {m \choose 1 }\cos^{m-1}a \sin\, a - {m \choose 3 }\cos^{m-3}a \sin ^3 a + ....  $
putting  m = 2n + 1 we get
$\sin(2n+1) a = {2n+ 1 \choose 1 }\cos^{2n}a \sin\, a - {2n+1 \choose 3 }\cos^{2n-2}a \sin ^3 a + ....  $
=  $\sin^{2n+1}a ({2n+ 1 \choose 1 }\cot^{2n}a - {2n+1 \choose 3 }\cot^{2n-2}a + ....) $
The LHS hand hence RHS  becomes zero for $a = \frac{k\pi}{2n+1}$
so for above t
${2n+ 1 \choose 1 }\cot^{2n}a - {2n+1 \choose 3 }\cot^{2n-2}a +... = 0 $
so $\cot^2\frac{\pi}{2n+1}$, $\cot^2\frac{2\pi}{2n+1}$ ... $\cot^2\frac{n\pi}{2n+1}$
are roots of the equation
$({2n+ 1 \choose 1 }x^n - {2n+1 \choose 3 }x^{n-1}  + ....=0$
so sum of roots
=  $\sum\limits_{k=1}^n \cot^2\frac{k}{2n+1} =\frac{{2n+1 \choose 3 }}{{2n+1 \choose 1 }} $
= $\frac{n(2n-1)}{3}$