How do we find the sum of this series $$ \sum_{i=n}^m \binom{i}{r} $$ where, $r<n<m$
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1Related: Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n\binom{t}{k}=\binom{n+1}{k+1}$. – JMoravitz Dec 10 '19 at 14:07
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1Rewrite your sum as $\sum\limits_{i=n}^m\binom{i}{r} = \sum\limits_{i=0}^m\binom{i}{r} - \sum\limits_{i=0}^{n-1}\binom{i}{r}$ and apply the hockey stick identity twice to get $\binom{m+1}{r+1}-\binom{n}{r+1}$. – JMoravitz Dec 10 '19 at 14:07
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Thanks for helping. I was a expecting a boring long algebraic proof but this hockey stick identify was unexpectedly clever – Nick Dec 11 '19 at 15:18