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I am having problems with proving the following formula by using induction.

$${r \choose r} + {r+1 \choose r} + {r+2 \choose r} + ... + {n \choose r} = {n+1 \choose r+1}$$

It is stated in the question that the r is arbitrary but fixed, which I also don't understand. If anyone could guide my how to solve this I would be very grateful.

Djboboch
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  • This is called the hockey-stick identity. – N. F. Taussig Nov 30 '17 at 12:13
  • But what does r is arbitrary but fixed mean can some one explain – Djboboch Nov 30 '17 at 12:28
  • @Djboboch, "$r$ is arbitrary but fixed" means, in this case, that while $r$ is some arbitrary number, you consider only one $r$ for the entire proof; unlike $n$, over which you try to do induction. That is: fix $r$. Then consider the base case, namely $n = r$; then the equation states $\binom rr = \binom{r+1}{r+1}$, which is true. Now suppose that the equation is true for $n$; then prove it for $n + 1$. – Mees de Vries Nov 30 '17 at 12:34
  • Observes that ${j \choose r} = 0~~~j<r$ and this [This ][1] [1]:https://math.stackexchange.com/questions/1490794/proof-of-the-hockey-stick-identity-sum-limits-t-0n-binom-tk-binomn1 – Guy Fsone Nov 30 '17 at 12:40
  • In my opinion, "$r$ is fixed" is irrelevant for the statement of the identity itself, but it's essential for the proof. – Arthur Nov 30 '17 at 12:42
  • See alsp (https://math.stackexchange.com/q/972368) – Jean Marie Nov 30 '17 at 12:44

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