$\sum_{i=1}^{n-k}\frac{i}{n-k}\binom{2n-2k}{n-k+i}\frac{1}{2}^{2(n-k)}=\frac{1}{2}^{2n-2k}\binom{2(n-k)-1}{n-k}$

Question

2$\sum_{i=1}^{n-k}\frac{i}{n-k}\binom{2n-2k}{n-k+i}\frac{1}{2}^{2(n-k)}=2\frac{1}{2}^{2n-2k}\binom{2(n-k)-1}{n-k}$.

This is an identity in a note for a class in Markov Processes, but I can't understand how this inequality holds. How does this work? I would greatly appreciate any help.

Answer 1

We need to compute: $$S_N=\sum_{i=1}^{N}\binom{2N}{N+i}i \tag{1}$$ but: $$ \sum_{i=1}^{N}\binom{2N}{N+i}N = \frac{N}{2}\left(4^N-\binom{2N}{N}\right)\tag{2}$$ and: $$ \sum_{i=1}^{N}\binom{2N}{N+i}(N+i) = 2N\sum_{i=1}^{N}\binom{2N-1}{N-i}=\frac{N}{2}\cdot 4^N\tag{3}$$ from the symmetry of binomial coefficients and the binomial formula. By $(2)$ and $(3)$ we have:

$$S_N=\sum_{i=1}^{N}\binom{2N}{N+i}i = \frac{N}{2}\binom{2N}{N} = \color{red}{N\binom{2N-1}{N}}\tag{4}$$

as wanted.