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Fix two natural numbers $m$ and $n$ accordingly. Using Wolfram we have

$$\sum_{k=1}^m\binom{k+n}{n}=\dfrac{(m+1)\binom{m+n+1}{n}-n-1}{n+1}.$$

Question: Can someone give me either an appropriate reference in which this formula might have appeared in so I can cite the result immediately in a technical report I am writing or a hint how to prove the identity without taking so much space (or an intuitive combinatorial sketch obvious to any reader)? Thanks in advance.

Also given likely similar (or exact) sum have been asked before in stack exchange I would like to apologise in advance for this redundant posting (there are too many of them for me to check). Do give me the link to the relevant previous posting if you don't mind.

RobPratt
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Sampah
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    You are likely to find it in the neater form $$\sum_{k=0}^m\binom{k+n}n=\frac{m+1}{n+1}\binom{m+n+1}n$$ that is $$\sum_{k=0}^m\binom{k+n}n=\binom{m+n+1}{n+1}$$ – bof Aug 04 '20 at 01:40
  • @bof Sweet! Thanks a lot. I should have thought that much earlier. In this neater form a variation of standard combinatorial argument should be possible. – Sampah Aug 04 '20 at 01:47
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    Or better yet, setting $p=m+n$, $$\sum_{h=0}^p\binom hn=\binom{p+1}{n+1}$$ – bof Aug 04 '20 at 01:49
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    The so-called hockey stick identity https://en.wikipedia.org/wiki/Hockey-stick_identity – bof Aug 04 '20 at 01:50

1 Answers1

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It’s essentially the hockey stick identity.

$$\sum_{k=1}^m\binom{n+k}n=\sum_{k=1}^{m+n}\binom{k}n=\sum_{k=0}^{m+n}\binom{k}n-1\,,$$

since $\binom{k}n=0$ when $k<n$, and as you can see (with proofs) at the link, that last summation simplifies to $\binom{m+n+1}{n+1}$, giving you identity

$$\sum_{k=1}^m\binom{n+k}n=\binom{m+n+1}{n+1}-1\,.$$

(Note that the Wolfram result simplifies to this.)

Brian M. Scott
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