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Possible Duplicate:
Identity involving binomial coefficients

how to prove $$ \sum_{k=1}^{n}\cot^2\left( \frac{k\pi}{2n+1}\right)=\frac{n(2n-1)}{3} $$ another $$ \sum_{k=0}^n{\binom {2k} {n} \binom {2n-2k} {n-k}}=4^n $$

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    Is it intentional that the second sum contains zero terms for $k < \dfrac n 2$ or do you intend $\binom {2n} k$ in place of $\binom {2k} n$?2012-10-15
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    for the first question take a look at [this](http://math.stackexchange.com/questions/173447/proving-sum-limits-l-1n-sum-limits-k-1n-1-tan-frac-lk-pi-2n1-t/173649#173649)2012-10-15
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    and [this](http://math.stackexchange.com/a/2343/19538)2012-10-15
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    [This answer](http://math.stackexchange.com/a/72661/12042), as corrected a few months ago, gives a complete proof of the second identity. It’s a more understandable version of the proof in the Sved article in the *Math. Intelligencer* mentioned [here](http://math.stackexchange.com/a/37984/12042).2012-10-15

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