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How do I evaluate $\sum_{k=0}^{n} \frac{(m+k)!}{m!k!}$ for a given n and m.

I got to know $\sum_{k=0}^{n} \frac{(m+k)!}{m!k!} = \sum_{k=0}^{n}\binom{m+k}{k} = \binom{m+n+1}{n}$ by asking ChatGPT, but I don't know how to prove it. Any help is greatly appreciated.

Jianzhao Zhu
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    "I got to know ... by asking ChatGPT" That seems like a recipe for failure. ChatGPT does not understand mathematics... it just repeats things that sound like things it has heard. It happens to have been correct in this specific case... but I would not trust it as a matter of principle. – JMoravitz Apr 19 '23 at 11:32
  • The first equality follows from the definition of $\binom{n}{\ell} = \frac{n!}{(n-\ell)!\ell!}$; for the second, see, e.g., https://math.stackexchange.com/questions/341156/sum-of-binomial-coefficients-sum-k-0n-binomkmm?rq=1 (and the fact that $\binom{m+n+1}{n}=\binom{m+1+n}{m+1}$). – Clement C. Apr 19 '23 at 11:33

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