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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!

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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.

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

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