# Prove $\sum_{i=0}^n\binom{m+i}{i}=\binom{m+n+1}{n}$ (another Hockey-Stick Identity?) [closed]

Let $$n$$ be a nonnegative integer, and $$m$$ a positive integer. Could someone explain to me why the identity $$\sum_{i=0}^n\binom{m+i}{i}=\binom{m+n+1}{n}$$ holds?

## closed as off-topic by Hanul Jeon, Leucippus, José Carlos Santos, Eevee Trainer, Parcly TaxelMar 26 at 3:27

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• You may want to see this: math.stackexchange.com/questions/1784613/…. (Replace their $r$ with your $n$ and their $n$ with your $m+1$.) – Minus One-Twelfth Mar 22 at 3:17
• This is easy to show by induction ... try it ? – Donald Splutterwit Mar 22 at 3:22
• Alternative answer ... Yes someone can explain it ! – Donald Splutterwit Mar 22 at 3:23
• @DonaldSplutterwit I know that it's done by induction but still sure how. So yes, I've tried it. – Ben Mar 22 at 4:05
• @MinusOne-Twelfth It's almost impossible to keep the terms straight but I substituted my stuff over to the best reply on that post but it still isn't very clear. Do you know the induction method? – Ben Mar 22 at 4:06

• The color #ffd is so ease to work with :) – Le Anh Dung Mar 22 at 5:52
$$\sum_{q=0}^n {m+q\choose q} = \sum_{q\ge 0} {m+q\choose q} [[0\le q\le n]] \\ = \sum_{q\ge 0} {m+q\choose q} [z^n] \frac{z^q}{1-z} = [z^n] \frac{1}{1-z} \sum_{q\ge 0} {m+q\choose q} z^q \\= [z^n] \frac{1}{1-z} \frac{1}{(1-z)^{m+1}} = [z^n] \frac{1}{(1-z)^{m+2}} = {n+m+1\choose n}.$$