Prove that $\sum_{k=0}^{n-1}\frac{n-k}{n}{2n \choose k}=\frac{1}{2}{2n \choose n}$

Question

I have tried splitting this sum into $$\sum_{k=0}^{n-1}\frac{n-k}{n}{2n \choose k}=\sum_{k=0}^{n-1}{2n \choose k}-\frac{1}{n}\sum_{k=0}^{n-1}k{2n \choose k}=\frac{2^{2n}-{2n \choose n}}{2}-\frac{1}{n}\sum_{k=0}^{n-1}k{2n \choose k}$$ but I get stuck on this step. I feel there may be an easier way to prove the identity using a generating function but I'm not sure.

Answer 1

It's when k varies from 0 to 2n that you get $2^{2n}$

In the second sum use the formula: $k.{2n\choose k}=2n.{2n-1\choose k-1}$

You are going to get

$\sum_{k=0}^{n-1} {2n\choose k} - \frac{2n}{n}\sum_{k=1}^{n-1} {2n-1\choose k-1}$

$=1+\sum_{k=1}^{n-1} {2n\choose k} - 2\sum_{k=1}^{n-1} {2n-1\choose k-1}$

$=1+\sum_{k=1}^{n-1} \left({2n\choose k}-{2n-1\choose k-1}\right) - \sum_{k=1}^{n-1} {2n-1\choose k-1}$

Using Pascal's formula and changing variable in the second sum $k'=k-1$, we get

$1+\sum_{k=1}^{n-1} {2n-1\choose k} - \sum_{k'=0}^{n-2} {2n-1\choose k'}$

$={2n-1\choose n-1}=\frac12 {2n\choose n}$