How to mathematically prove: $\sum_{x = r}^{n} C(x , r) = C(n+1 , r+1)$.

Question

The RHS is equal to the number of ways of selecting $r+1$ out of $n+1$ objects. If I sum the number of selections when the $(r+1)$th (i.e. the last) selection is the at the $(x+1)$th position, I get the RHS.

I verified it with some values but couldn't prove it mathematically.

Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. — N. F. Taussig, Sep 28 '18 at 10:10
Check out various proofs here: https://en.wikipedia.org/wiki/Hockey-stick_identity — farruhota, Sep 28 '18 at 10:18
Possible duplicate of Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$ — Ѕᴀᴀᴅ, Sep 28 '18 at 10:22

score 1 · Answer 1 · answered Sep 28 '18 at 10:09

1

The usual notation is $\sum_{x=r}^n\binom{x}{r}=\binom{n+1}{r+1}$. Since $\binom{x}{r}=\binom{x+1}{r+1}-\binom{x}{r+1}$, the sum telescopes to $\binom{n+1}{r+1}-\binom{r}{r+1}$. But the last term is, of course, $0$.

answered Sep 28 '18 at 10:09

J.G.

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How to mathematically prove: $\sum_{x = r}^{n} C(x , r) = C(n+1 , r+1)$.

1 Answers1