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In my book, it claimed $\binom{2}{2} + \binom{3}{2} + \binom{4}{2} + ... + \binom{9}{2} + \binom{10}{2} + \binom{11}{2} = \binom{12}{3}$.

I can't figure out How the author claimed the above true. Is there any method without just calculating?

Plus, From above I guess the $\binom{n}{2} + \binom{n+1}{2} = \binom{n+1}{3}$. Then, Does $\binom{n}{r} + \binom{n+1}{r} = \binom{n+1}{r+1}$ hold in general?

kechang lee
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    Try $r=1$. Do you get equality in this case? $\binom{n}{2} + \binom{n+1}{2} = \binom{n+1}{3}$ is also false. – Kavi Rama Murthy Apr 07 '22 at 08:43
  • @KaviRamaMurthy, It didn't occur to me to think the $r=1$ case. Yes. I got it. It is not true in general. – kechang lee Apr 07 '22 at 08:49
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    The sum is just hockey-stick identity – Sil Apr 07 '22 at 08:51
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    Re the comment of @Sil, you can tell at a glance that the hockey stick identity is valid, without doing (much) Math, simply by using the Pascal's Triangle rule that each number is the sum of the 2 numbers above it. When applying this rule against (for example) the Wikipedia diagram in the article cited by Sil's comment, simply put a crook into the top of the hockey stick, substituting $~\displaystyle \binom{3}{0} = 1~$ for $~\displaystyle \binom{2}{0} = 1.$ – user2661923 Apr 07 '22 at 11:15

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