Find sum binomial coefficients $$
\sum_{k>=1}^{\infty} {2N \choose N-k}k
$$
How to find this sum?
I know that the answer is $ \frac{1}{2}N{2N \choose N}$
But it is very interesting to know the solution :)
 A: Our first goal in dealing with this sum is to get rid of the $k$. The standard approach is to rewrite something like $\binom{n}{k} \cdot k$ as $\frac nk \binom{n-1}{k-1} \cdot k$, or $n \binom{n-1}{k-1}$. Here, the bottom index doesn't match the extra factor, but we can make it so with a little extra work:
\begin{align}
  \binom{2N}{N-k} k &= \binom{2N}{N+k}k \\
    &= \binom{2N}{N+k}(N+k) - \binom{2N}{N+k} N \\
    &= \frac{2N}{N+k} \binom{2N-1}{N+k-1}(N+k) - N \binom{2N}{N+k} \\
    &= 2N \binom{2N-1}{N+k-1} - N \binom{2N}{N+k}.
\end{align}
At this point, we have two sums that are both easier to deal with:
$$\sum_{k \ge 1} \binom{2N}{N-k} k = 2N \sum_{k \ge 1} \binom{2N-1}{N+k-1} - N \sum_{k \ge 1} \binom{2N}{N+k}.$$
The sum of all binomial coefficients of the form $\binom{2N-1}{i}$ is $2^{2N-1}$, and our first sum takes only those binomial coefficients of this form where $i \ge N$. These are the second half, which by symmetry is equal to the first half, so the first sum simplifies to $2^{2N-2}$.
We're in much the same position with the second sum, except that the $\binom{2N}{i}$ coefficients also have a central coefficient $\binom{2N}{N}$, which is left out here. The sum of all coefficients that aren't the central one is $2^{2N} - \binom{2N}{N}$, and this sum is half of that.
Putting these facts together, we get
\begin{align}
\sum_{k \ge 1} \binom{2N}{N-k} k 
  &= 2N \Bigg(2^{2N-2}\Bigg) - N \Bigg(2^{2N-1} - \frac12\binom{2N}{N}\Bigg) \\
  &= N \cdot 2^{2N-1} - N \cdot 2^{2N-1} + \frac N2 \cdot \binom{2N}{N} \\
  &= \frac N2 \cdot \binom{2N}{N}.
\end{align}
A: Here is a slightly different variation.

We obtain
  \begin{align*}
\sum_{k=1}^\infty\binom{2N}{N-k}k&=\sum_{k=0}^N\binom{2N}{N-k}k\tag{1}\\
&=\sum_{k=0}^N\binom{2N}{k}(N-k)\tag{2}\\
&=N\sum_{k=0}^N\binom{2N}{k}-\sum_{k=1}^N\binom{2N}{k}k\\
&=N\left(2^{2N-1}+\frac{1}{2}\binom{2N}{N}\right)-2N\sum_{k=1}^N\binom{2N-1}{k-1}\tag{3}\\
&=2^{2N-1}N+\frac{N}{2}\binom{2N}{N}-2N\sum_{k=0}^{N-1}\binom{2N-1}{k}\tag{4}\\
&=2^{2N-1}N+\frac{N}{2}\binom{2N}{N}-2N\cdot2^{2N-2}\tag{5}\\
&=\frac{N}{2}\binom{2N}{N}
\end{align*}

Comment:


*

*In (1) we set the upper limit of the sum to $N$ since $\binom{2N}{N-k}=0$ for $k>N$. We also start with $k=0$ without changing anything since we are adding zero only.

*In (2) we change the order of summation $k\rightarrow N-k$.

*In (3) we use the symmetry of the binomial coefficients and the fact that they sum up to $2^{2N}$. We have additionally to respect the central binomial coefficient since  we have an odd number of summands. We also use the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$ for the right-hand sum.

*In (4) we shift the index of the sum by $1$.

*In (5) we  again use  the symmetry of the binomial coefficients and the fact that they sum up to $2^{2N-1}$. This time we have an even number of summands and need not to care for a central binomial coefficient.
