$1,3,6,10,15,...$ 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.
