Find $\sum_{k=1}^n \binom{2n-k}{n}(-1)^k$

Question

Is there closed form for $\sum_{k=1}^n \binom{2n-k}{n}(-1)^k$?

I got above expression for a counting exercise. I wonder that it might have the closed form but I am not sure yet. Can anyone have any idea? Thank you in advance !

Answer 1

$$\sum_{k=0}^{n}(-1)^k\binom{2n-k}{k} = \binom{2n}{n}\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{k!(2n-k)!}{(2n)!} $$ gives: $$\sum_{k=0}^{n}(-1)^k\binom{2n-k}{k} = (2n+1)\binom{2n}{n}\sum_{k=0}^{n}\binom{n}{k}(-1)^k \int_{0}^{1} x^{2n-k}(1-x)^k\,dx $$ hence: $$\sum_{k=0}^{n}(-1)^k\binom{2n-k}{k} = (2n+1)\binom{2n}{n}\int_{0}^{1} x^{2n}\left(1+\frac{x-1}{x}\right)^n\,dx $$ or:

$$\sum_{k=0}^{n}(-1)^k\binom{2n-k}{k} = \color{red}{(2n+1)\binom{2n}{n}\int_{0}^{1} x^n(2x-1)^n\,dx}. $$

The RHS is a hypergeometric function, with a nice integral representation.