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}

Let $$S_n=\sum_{k=0}^{n} {2n \choose k}$$
and let
$$S'_n=\sum_{k=0}^{n} k {2n \choose n-k}=\sum_{k=0}^{n} (n-k) {2n \choose k}=nS_n-T_n$$
$$2^{2n}=\sum_{k=0}^{2n} {2n \choose k}=\sum_{k=0}^{n} {2n \choose k}+\sum_{k=n+1}^{2n} {2n \choose k}$$
$$\implies S_n+\sum_{k=n+1}^{2n} {2n \choose k}=2^{2n}$$
Change $k$ to $2n-j$, then
$$S_n+\sum_{j=n-1}^{0} {2n \choose 2n-j}=2^{2n}$$
$$\implies S_n+ \sum_{j=0}^{n} {2n \choose j}-{2n \choose n}=2^{2n}$$
$$\implies S_n=2^{2n-1}+\frac{1}{2}{2n \choose n}.$$
Next let
$$T_n=\sum_{k=0}^{n} k {2n \choose k}=\sum_{k=0}^{n} k\frac{(2n)!}{k!(2n-k)!}=2n\sum_{k=0}^{n}{2n-1 \choose k-1}=2n\sum_{j=0}^{n-1}{2n-1 \choose j} $$
We have $$2^{2n-1}= \sum_{k=0}^{2n-1} {2n-1\choose k}=\sum_{k=0}^{n-1} {2n-1 \choose k}+\sum_{k=n}^{2n-1} {2n-1 \choose k}=\sum_{k=0}^{n-1} {2n-1 \choose k}+\sum_{j=0}^{n-1} {2n-1 \choose n+j}$$ $$2^{2n-1}\implies \sum_{k=0}^{n-1} {2n-1 \choose k}+\sum_{j=0}^{n-1} {2n-1 \choose n+j}=\sum_{k=0}^{n-1} {2n-1 \choose k}+\sum_{j=0}^{n-1} {2n-1 \choose n-1-j}$$ $$\implies 2^{2n-1}= \sum_{k=0}^{n-1} {2n-1 \choose k}+\sum_{m=0}^{n-1} {2n-1 \choose m} \implies \sum_{k=0}^{n-1} {2n-1 \choose k}=2^{2n-2}$$
Finally, we have $$S'_n=n2^{2n-1}+\frac{n}{2} {2n \choose n}-n2^{2n-1}=\frac{n}{2}{2n \choose n}$$
