Prove that $\sum\limits_{i=1}^n 2i\binom{2n}{n-i}= n\binom{2n}{n}$ Let n be a positive integer.

Prove that $$\sum_{i=1}^n 2i\binom{2n}{n-i}= n\binom{2n}{n}.$$

 A: \begin{align}
\sum_{i=1}^{n}{2i\binom{2n}{n-i}} & =\sum_{i=1}^{n}{(n+i)\binom{2n}{n-i}}-\sum_{i=1}^{n-1}{(n-i)\binom{2n}{n-i}} \\
& =\sum_{i=1}^{n}{2n\binom{2n-1}{n-i}}-\sum_{i=1}^{n-1}{2n\binom{2n-1}{n-i-1}} \\
& =\sum_{i=0}^{n-1}{2n\binom{2n-1}{i}}-\sum_{i=0}^{n-2}{2n\binom{2n-1}{i}} \\
& =2n\binom{2n-1}{n-1} \\
& =n \binom{2n}{n}
\end{align}
A: Here’s an alternative that requires a little less of a leap to get started, but a little more algebra later. Instead of pulling $2i=(n+i)-(n-i)$ out of thin air, substitute $k=n-i$:
$$\begin{align*}
\sum_{i=0}^n2i\binom{2n}{n-i}&=\sum_{k=0}^{n-1}2(n-k)\binom{2n}k\\
&=2n\sum_{k=0}^{n-1}\binom{2n}k-2\sum_{k=0}^{n-1}k\binom{2n}k\\
&=2n\sum_{k=0}^{n-1}\binom{2n}k-4n\sum_{k=1}^{n-1}\binom{2n-1}{k-1}\\
&=2n\left(\sum_{k=0}^{n-1}\binom{2n}k-2\sum_{k=0}^{n-2}\binom{2n-1}k\right)\\
&=2n\left(\sum_{k=0}^{n-1}\binom{2n}k-\sum_{k=0}^{n-2}\left(\binom{2n-1}k+\binom{2n-1}{2n-1-k}\right)\right)\\
&=2n\left(\sum_{k=0}^{n-1}\binom{2n}k-\sum_{k=0}^{n-2}\binom{2n-1}k-\sum_{k=n+1}^{2n-1}\binom{2n-1}k\right)\\
&=2n\left(\sum_{k=0}^{n-1}\binom{2n}k-\left(2^{2n-1}-\binom{2n-1}{n-1}-\binom{2n-1}n\right)\right)\\
&=2n\left(\sum_{k=0}^{n-1}\binom{2n}k-2^{2n-1}+\binom{2n}n\right)\\
&=n\left(2\sum_{k=0}^{n-1}\binom{2n}k-2^{2n}+2\binom{2n}n\right)\\
&=n\left(2^{2n}-\binom{2n}n-2^{2n}+2\binom{2n}n\right)\\
&=n\binom{2n}n\;.
\end{align*}$$
I still haven’t found a combinatorial argument, though.
