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I'm on an exercise for the university and have no clue what to do. The exercise is the following: You should proof that $\sum_{i=k}^n {i \choose k}= {{n+1}\choose {k+1}}$ . I've done that with an induction over $n$ and this goes very well.

But the other part is that we should use another formula an that is the following to proof it:

$$\sum_{i=k}^n {{i+1}\choose{k+1}} = \sum_{i=k}^n {i \choose{k+1}} + {i \choose{k}}$$

And there i need your help. Does anyone know what to do with this??

Thanks for your help!

Elisabeth Pachl
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  • As for the content of what appears to actually be your question, $\binom{i+1}{k+1}=\binom{i}{k+1}+\binom{i}{k}$ is just pascal's identity. Rewritten, $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$, if you want to pick $r$ people from $n$ where one person is considered special, you may either pick the special person and $r-1$ additional people from the $n-1$ non-special people or you may choose not to pick the special person and instead pick all $r$ people from the $n-1$ not-special people – JMoravitz May 04 '17 at 18:58
  • Once the statement is corrected, I expect that the identity you are attempting to prove is the hockey stick identity – JMoravitz May 04 '17 at 19:01
  • i will look that up! the first look seems very promising! thanks! – Elisabeth Pachl May 04 '17 at 19:03

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