Prove

\[\displaystyle \sum_{k=0}^n{\binom{n}{k} \binom{2k}{k} {\left(\frac{-1}{2}\right)}^k} = \begin{cases} \frac{1}{2^n}\binom{n}{n/2}, & \text{n is even} \\ 0, & \text{n is odd} \end{cases}\]

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## Comments

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TopNewestSince \[ {2k \choose k} \; = \; \frac{1}{2\pi}\int_{-\pi}^\pi e^{-ikt}(1 + e^{it})^{2k}\,dt \; = \; \frac{1}{2\pi}\int_{-\pi}^\pi \big(2\cos\tfrac12kt\big)^{2k}\,dt \] we have \[\begin{align} \sum_{k=0}^n {n \choose k}{2k \choose k}\big(-\tfrac12\big)^k & = \frac{1}{2\pi}\int_{-\pi}^\pi {n \choose k}\big(2\cos\tfrac12kt\big)^{2k} \big(-\tfrac12\big)^k\,dt \\ & = \frac{1}{2\pi}\int_{-\pi}^\pi \Big(1 - 2\cos^2\tfrac12kt\Big)^n\,dt \; = \; \frac{(-1)^n}{2\pi}\int_{-\pi}^\pi \cos^nt\,dt \\ & = \frac{(-1)^n}{2^{n+1}\pi}\int_{-\pi}^\pi \big(e^{it} + e^{-it}\big)^n\,dt \; = \; \frac{(-1)^n}{2^{n+1}\pi}\sum_{k=0}^n {n \choose k} \int_{-\pi}^{\pi} e^{i(2k-n)t}\,dt \\ & = \left\{ \begin{array}{lll} \frac1{2^n}{n \choose \frac12n} & \hspace{1cm} & n \mbox{ even} \\ 0 & & n \mbox{ odd} \end{array} \right. \end{align} \]

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@Mark Hennings

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