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How would you proof $$ \sum_{m=k}^{n}\binom{m}{k} = \binom{n + 1}{k + 1} $$ with $n \geq k$ and $n$, $k \in \mathbb{N}$ by induction? I had some approaches but wasn't sure if they were right, so I'd appreciate if you could share some solutions!

Przemysław Scherwentke
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user2103480
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  • Why don't you show us your induction proof so we can see if it is correct or not? The proof is not hard at all using induction. – rogerl Oct 29 '14 at 22:19
  • possible duplicate: http://math.stackexchange.com/questions/398771/proving-sum-m-0m-binommkk-binomkm1k1 – robjohn Oct 29 '14 at 22:46

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The induction proof on $n$ reduces to Pascal's identity $$\binom{n+1}{k+1} + \binom{n+1}{k} = \binom{n+2}{k+1}.$$ But there is a nicer combinatorial proof of your identity. The right-hand side describes the number of ways of choosing a subset $S$ of $\{1,\ldots,n+1\}$ of size $k+1$. Let $m = \max S - 1$, so $k \leq m \leq n$. Then $S \setminus \{m+1\}$ is a subset of $\{1,\ldots,m\}$ of size $k$. This is what the left-hand side describes.

Yuval Filmus
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  • Does this really suffice? If we used $n=0$ and $k=0$ as the induction's base, wouldn't this just prove the theorem for any n and $k=0$? – user2103480 Oct 29 '14 at 23:00
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    The induction is only on $n$. For any $k$ you have a base case $n = k$ which reads $\binom{k}{k} = \binom{k+1}{k+1}$. – Yuval Filmus Oct 29 '14 at 23:48