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Question: $$\sum_{k=1}^{n}k\binom{2n}{n+k}=\frac n2\binom{2n}{n}$$ is true for every $n\in \mathbb N$? If this is true, then how can we prove this?

When I was playing with numbers, I conjectured this. Though I found that this holds for many $n$s, I haven't found any proof. Can anyone help?

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    $\begingroup$ You are really good at playing with numbers $\endgroup$ – evil999man May 12 '14 at 16:38
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Let $S$ be the considered sum we have $$\eqalign{ S&=\sum_{k=1}^nk\binom{2n}{n+k}=\sum_{k=n+1}^{2n}(k-n)\binom{2n}{k}\cr &=\sum_{k=n+1}^{2n}k\binom{2n}{k}-n\sum_{k=n+1}^{2n} \binom{2n}{k}\cr &=2n\sum_{k=n+1}^{2n} \binom{2n-1}{k-1}-n\sum_{k=n+1}^{2n} \binom{2n}{k}\cr &=n\sum_{k=n+1}^{2n} \left(2\binom{2n-1}{k-1}- \binom{2n}{k}\right)\cr &=n\sum_{k=n+1}^{2n} \left(2\binom{2n-1}{k-1}- \binom{2n-1}{k}- \binom{2n-1}{k-1}\right)\cr &=n\sum_{k=n+1}^{2n} \left(\binom{2n-1}{k-1}- \binom{2n-1}{k}\right)\cr &=n\binom{2n-1}{n}=\frac{n}{2}\binom{2n}{n}. }$$ which the desired conclusion.$\qquad\square$

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$$\eqalign{\sum_{k=1}^n k\binom{2n}{n+k}&=\sum_{k=1}^n (n+k)\binom{2n}{n+k}-n\sum_{k=1}^n\binom{2n}{n+k}\cr &=\sum_{k=n+1}^{2n} k\binom{2n}{k}-\frac{n}{2}(\sum_{k=0}^{2n}\binom{2n}{k}-\binom{2n}{n})\cr&=2n\sum_{k=n+1}^{2n}\binom{2n-1}{k-1}-\frac{n}{2}(2^{2n})+\frac{n}{2}\binom{2n}{n}\cr&=n\sum_{k=0}^{2n-1}\binom{2n-1}{k}-\frac{n}{2}(2^{2n})+\frac{n}{2}\binom{2n}{n}\cr&=\frac{n}{2}\binom{2n}{n} }$$

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