show that; strange sum yields triangular numbers $\sum_{k=1}^{n}\tan^2\left({k\pi\over 2n+1}\right)=T_{2n}$ $1,3,6,10,15,...$ for $n=1,2,3,...$ it is the n-th Triangular numbers.
I find it unsual that this sum yields even triangular numbers;
$$\sum_{k=1}^{n}\tan^2\left({k\pi\over 2n+1}\right)=T_{2n}$$
How can I show that? Any hints into this strange sum?
I can't figure it out where to start!

As recommended in the comment section of OPs question we can follow the nice answer of Fallager.

We obtain
  \begin{align*}
\left(\cos\frac{k\pi}{2n+1}+i\sin\frac{k\pi}{2n+1}\right)^{2n+1}&=(-1)^k\tag{1}\\
\\
\sum_{j=0}^{2n+1}\binom{2n+1}{j}\left(\cos\frac{k\pi}{2n+1}\right)^j\left(i\sin \frac{k\pi}{2n+1}\right)^{2n+1-j}&=(-1)^k\tag{2}\\
\\
\sum_{j=0}^{n}\binom{2n+1}{2j}\left(\cos\frac{k\pi}{2n+1}\right)^{2j}\left(i\sin \frac{k\pi}{2n+1}\right)^{2n+1-2j}&=0\tag{3}\\
\\
\sum_{j=0}^{n}\binom{2n+1}{2j}\left(i\tan \frac{k\pi}{2n+1}\right)^{2n+1-2j}&=0\tag{4}\\
\\
\sum_{j=0}^{n}\binom{2n+1}{2j}\left(i\tan \frac{k\pi}{2n+1}\right)^{2n-2j}&=0\tag{5}\\
\end{align*}

Comment:


*

*In (1) we use De Moivre's formula

*In (2) we apply the Binomial theorem

*In (3) we take the imaginary part of (2) i.e. with index $j$ even

*In (4) we divide by $\left(\cos \frac{k\pi}{2n+1}\right)^{2n+1}$

*In (5) we divide by $\tan \frac{k\pi}{2n+1}$

We conclude $\left(\tan\frac{k\pi}{2n+1}\right)^2$ with $0\leq k\leq 2n$ are the zeros of the following polynomial in $z$
  \begin{align*}
\sum_{j=0}^n\binom{2n+1}{2j}\left(-z\right)^{n-j}
\end{align*}
According to Vieta's formulas the sum of the zeros is
  \begin{align*}
  \sum_{k=0}^{n}\left(\tan\frac{k\pi}{2n+1}\right)^2=\frac{\binom{2n+1}{2}}{\binom{2n+1}{0}}=n(2n+1)=T_{2n}
  \end{align*}
  and the claim follows.

