TITLE: A conjecture including binomial coefficients
QUESTION [16 upvotes]: 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?

REPLY [7 votes]: $$\eqalign{\sum_{k=1}^n k\binom{2n}{n+k}&=\sum_{k=1}^n (n+k)\binom{2n}{n+k}-n\sum_{k=1}^n\binom{2n}{n+k}\cr
&=\sum_{k=n+1}^{2n} k\binom{2n}{k}-\frac{n}{2}(\sum_{k=0}^{2n}\binom{2n}{k}-\binom{2n}{n})\cr&=2n\sum_{k=n+1}^{2n}\binom{2n-1}{k-1}-\frac{n}{2}(2^{2n})+\frac{n}{2}\binom{2n}{n}\cr&=n\sum_{k=0}^{2n-1}\binom{2n-1}{k}-\frac{n}{2}(2^{2n})+\frac{n}{2}\binom{2n}{n}\cr&=\frac{n}{2}\binom{2n}{n}
}$$