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I'm looking for a proof of this identity but where j=m not j=0

http://www.proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Upper_Index

$$\sum_{j=m}^n\binom{j}{m}=\binom{n+1}{m+1}$$

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marked as duplicate by user147263, Davide Giraudo, Najib Idrissi, Aaron Maroja, Jack D'Aurizio Apr 8 '15 at 15:56

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    $\begingroup$ To clarify what you mean: you want a proof of $$\sum_{j=m}^n\binom{j}{m}=\binom{n+1}{m+1}$$ correct? $\endgroup$ – Zev Chonoles Oct 22 '11 at 15:02
  • $\begingroup$ (-1) ${}{}{}{}{}{}$ $\endgroup$ – The Chaz 2.0 Dec 21 '11 at 16:14
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It seems to me that you are looking for the proof of the identity: $$ \sum_{j=m}^n\binom{j}{m}=\binom{n+1}{m+1}$$

This is actually known as the Hockey-Stick Identity.You can find different methods of proving this in this page.

Please note that there are many applications of the hockey stick identity and all the combinatorial identities you may encounter, this is the one most worth remembering. It makes a lot of apparently hard problems very easy.

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    $\begingroup$ Any idea why it's called the hockey stick identity? $\endgroup$ – Alex Oct 22 '11 at 16:00
  • $\begingroup$ @Alex: when you go to the URL FoolForMath provided, it shows a picture with the above identity represented on Pascal's triangle, on which the relevant numbers form a 'hockey stick'. $\endgroup$ – Gerben Oct 22 '11 at 16:23
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    $\begingroup$ @Gerben: Oh, I guess that's a little hockey stick like. Neat! Thanks! $\endgroup$ – Alex Oct 22 '11 at 16:28
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Hint: what is $\binom{2}{4}$? what is $\binom{n-1}{n}?$

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