# Proof for a certain binomial identity

Let $$1\leq m\leq n$$ be positive integers. I will appreciate any help proving the following identity $$\sum_{k=n}^{n+m}(-1)^k\binom{k-1}{n-1}\binom{n}{k-m}=0$$ Thanks!

Here we have Chu-Vandermonde's Identity in disguise.

We obtain for $$1\leq m\leq n$$ \begin{align*} \color{blue}{\sum_{k=n}^{n+m}}&\color{blue}{(-1)^k\binom{k-1}{n-1}\binom{n}{k-m}}\\ &=\sum_{k=0}^m(-1)^{k+n}\binom{n+k-1}{n-1}\binom{n}{k+n-m}\tag{1}\\ &=\sum_{k=0}^m(-1)^{k+n}\binom{n+k-1}{k}\binom{n}{m-k}\tag{2}\\ &=(-1)^n\sum_{k=0}^m\binom{-n}{k}\binom{n}{m-k}\tag{3}\\ &=(-1)^n\binom{0}{m}\tag{4}\\ &\,\,\color{blue}{=0} \end{align*}

Comment:

• In (1) we shift the index to start with $$k=0$$.

• In (2) we apply the binomial identity $$\binom{p}{q}=\binom{p}{p-q}$$ twice.

• In (3) we apply the binomial identity $$\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$$.

• In (4) we finally apply the Chu-Vandermonde identity.

• Thanks Markus! It is not so trivial. – boaz Feb 24 at 15:56
• @boaz: You're welcome. – Markus Scheuer Feb 24 at 15:57