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 :)

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}
