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}

I don't konw how to prove it, but for general case I found something:
Integral Representation and the Computation of Combinatorial Sums by G.P.EGORYCHEV  Page77  
says that:


verify the formula$\sum\limits_{i=1}^n i\left(\begin{array}{c}
2 n \\
n-i
\end{array}\right)=\frac{1}{2}(n+1)\left(\begin{array}{c}
2 n \\
n-1
\end{array}\right)$ with Mathematica
Sum[Binomial[i, 1]*Binomial[2*n, n - i], {i, 1, n}]

$$
\frac{1}{2} (n+1) \binom{2 n}{n-1}
$$
