Binomial Coefficient Challenge 7!

Prove

∑k=0n(nk)(2kk)(−12)k={12n(nn/2),n is even0,n is odd\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}

Note by Kartik Sharma
4 years ago

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Comments

Since (2kk)  =  12π∫−ππe−ikt(1+eit)2k dt  =  12π∫−ππ(2cos⁡12kt)2k dt {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 ∑k=0n(nk)(2kk)(−12)k=12π∫−ππ(nk)(2cos⁡12kt)2k(−12)k dt=12π∫−ππ(1−2cos⁡212kt)n dt  =  (−1)n2π∫−ππcos⁡nt dt=(−1)n2n+1π∫−ππ(eit+e−it)n dt  =  (−1)n2n+1π∑k=0n(nk)∫−ππei(2k−n)t dt={12n(n12n)n even0n odd\begin{aligned} \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{aligned}

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Mark Hennings - 4 years ago

@Mark Hennings

Kartik Sharma - 4 years ago
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