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While calculating an integral in a quantum mechanical problem by two different methods, I came across the following identity $$\sum_{k=0}^n\sum_{m=0}^{2k}(-2)^m\binom{2(n-k)}{n-k}\binom{2k}{k}\binom{...

Let $p(x)=\sum_{k=1}^m [a_k\cos(n_kx)+b_k\sin(n_kx)]$ be a null average trigonometric polynomial (null average means that is $\int_\mathbb T p =0$ or, equivalently, there are no $a_0$ and $b_0$). ...