<s>[INST] Prove that $\sum_{k=1}^nk^2{n\choose k}^2=n^2 \binom {2n-2}{n-1}$ Please help me / give a hand with combinational prove for:
$$ 1^2 \binom  n 1 ^2 + 2^2 \binom n 2 ^2 + \dots + n^2 \binom n n ^2 = n^2 \binom {2n-2}{n-1}$$
 [/INST] We have:
$$\sum_{k=1}^{n}k\binom{n}{k}z^k = nz(1+z)^{n-1}\tag{1}$$
hence, replacing $z$ with $e^{\pm i\theta}$ and using the Parseval identity:
$$\begin{eqnarray*}\sum_{k=1}^{n}k^2\binom{n}{k}^2&=&\frac{n^2}{2\pi}\int_{0}^{2\pi}(1+e^{i\theta})^{n-1}(1+e^{-i\theta})^{n-1}\,d\theta\\&=&\frac{n^2}{2\pi}\int_{0}^{2\pi}(2+2\cos\theta)^{n-1}d\theta\\&=&\frac{n^2 4^{n}}{2\pi}\int_{0}^{\pi/2}\cos^{2n-2}(\theta)\,d\theta\\&=&n^2\binom{2n-2}{n-1}.\end{eqnarray*}$$
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