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}
 A: \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}
A: 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 $$S'_n=n2^{2n-1}+\frac{n}{2} {2n \choose n}-n2^{2n-1}=\frac{n}{2}{2n \choose n}$$
A: Related Equation
Just for a fun alternative, we have the following equation that we will prove below.
$$
\sum_{i=2}^{n} (n+i)\binom{2n}{n+i} -
\sum_{i=1}^{n} (n-i)\binom{2n}{n-i}
=
0
$$
Combinatorial Proof
One wish to colour $2n$ distinguishable balls such that exactly one is pink, some (can be none) are black, and some (can be none) are white. The two terms on LHS are, respectively:

*

*Number of possible colouring such that there are at least $n+1$ black balls

*Number of possible colouring such that there are at most $n-2$ white balls

Easy to see that (1) equals (2) so their difference is zero.
Relation to Original Question
$$
\begin{align}
\sum_{i=1}^{n} i\binom{2n}{n-i}
&=
\frac{1}{2}
\left[
\sum_{i=1}^{n} (n+i)\binom{2n}{n+i} -
\sum_{i=1}^{n} (n-i)\binom{2n}{n-i}
\right]
\\\\
&=
\frac{n+1}{2}\binom{2n}{n+1}
\end{align}
$$
Potential Extension
Instead of one pink, colour $m$ balls pink.
$$
\sum_{i=m+1}^{n}\binom{n+i}{m}\binom{2n}{n+i} -
\sum_{i=1}^{n}\binom{n-i}{m}\binom{2n}{n-i} =
0
$$
The two terms are, respectively:

*

*Number of possible colouring such that there are at least $n+1$ black balls.

*Number of possible colouring such that there are at most $n-m-1$ white balls.

These two are the same, thus the difference is zero and we can write:
$$
\sum_{i=1}^{n}\left[\binom{n+i}{m}-\binom{n-i}{m}\right]\binom{2n}{n-i}=
\sum_{i=1}^{m}\binom{n+i}{m}\binom{2n}{n-i}
$$
For odd $m$, $\binom{n+i}{m}-\binom{n-i}{m}$ has the term $i^{m}$
A: 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}]

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