Prove a binomial identity: $\sum_{i=1}^n i \binom{2n}{n-i}=\frac12(n+1) \binom{2n}{n-1}$ I want to prove the product of even/odd power with a combinatorial number:
\begin{aligned}
  \sum_{i=1}^n i \binom{2n}{n-i}&=\frac12(n+1) \binom{2n}{n-1},  \\
  \sum_{i=1}^n i^2 \binom{2n}{n-i}&=2^{2n-2} n.
\end{aligned}
I am sure these results are correct since WolframAlpha verifies them. I wonder if the first formula can be proved, and also the general version can be proved: Formula related to combinatorial number.

Below is proof for the second formula:
First, note that $r\binom{n}{r}=n\binom{n-1}{r-1}$, and $\sum_{i=0}^n\binom{n}{i}=2^n$. We thus have
\begin{aligned}
    \sum_{i=0}^n i\binom{n}{i}&=2^{n-1}n,  \\
    \sum_{i=0}^n i^2\binom{n}{i}
    &= \sum_{i=0}^n i(i-1)\binom{n}{i}+\sum_{i=0}^n i\binom{n}{i} \\
    &= 2^{n-2}(n^2-n) + 2^{n-1} n  \\
    &= n(n+1)2^{n-2}. 
\end{aligned}
Therefore,
\begin{aligned}
    \sum_{i=0}^{2n}(n-i)^2\binom{2n}{i}
    &= n^2\sum_{i=0}^{2n}\binom{2n}{i}-2n\sum_{i=0}^{2n}i\binom{2n}{i}+\sum_{i=0}^{2n}i^2 \binom{2n}{i}  \\
    &= 2^{2n}n^2-2^{2n+1}n^2+n(2n+1)2^{2n-1}  \\
    &= 2^{2n-1} n,
\end{aligned}
and thus
\begin{aligned}
    \sum_{i=0}^n i^2 \binom{2n}{n-i} 
    = \sum_{i=0}^n (n-i)^2 \binom{2n}{i} 
    = \frac12 \sum_{i=0}^{2n} (n-i)^2 \binom{2n}{i}
    = 2^{2n-2} n.
\end{aligned}

\begin{align}
\sum_{i=1}^n i\binom{2n}{n-i}
&= \sum_{i=0}^n i\binom{2n}{n-i} \\
&= \sum_{i=0}^n (n-i)\binom{2n}{i} \\
&= n\sum_{i=0}^n \binom{2n}{i} - \sum_{i=1}^n i\binom{2n}{i} \\
&= \frac{n}{2}\left(\binom{2n}{n}+\sum_{i=0}^{2n} \binom{2n}{i}\right) - 2n\sum_{i=1}^n \binom{2n-1}{i-1} \\
&= \frac{n}{2}\left(\binom{2n}{n}+2^{2n}\right) - 2n\sum_{i=0}^{n-1}\binom{2n-1}{i} \\
&= \frac{n}{2}\left(\binom{2n}{n}+2^{2n}\right) - \frac{2n}{2}\sum_{i=0}^{2n-1}\binom{2n-1}{i} \\
&= \frac{n}{2}\left(\binom{2n}{n}+2^{2n}\right) - n2^{2n-1}\\
&= \frac{n}{2}\binom{2n}{n}\\
&= \frac{n+1}{2}\binom{2n}{n-1}
\end{align}
