I was looking at some questions in a Cambridge text and I reached this question however I am at it for 1 hr and can't seem to get the proof right. Any help ?
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Eric Wofsey
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user122343
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Did you try googling it? https://en.wikipedia.org/wiki/Hockey-stick_identity – pancini Aug 18 '18 at 07:47
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For a proof see here. – drhab Aug 18 '18 at 08:13
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You can check Simplify the expression $\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots +\binom{n+k}{k}$ or Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$ and other posts linked there. – Martin Sleziak Aug 21 '18 at 02:08
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Assume it true for $p-1$, i.e. $${}^nC_0+{}^{n+1}C_1+{}^{n+2}C_2+\dots+{}^{n+p-1}C_{p-1}={}^{n+p}C_{p-1}$$ Let's prove it for $p$, i.e. $${}^nC_0+{}^{n+1}C_1+{}^{n+2}C_2+\dots+{}^{n+p}C_p={}^{n+p+1}C_p$$ So the left hand side is composed of the assumption in the first equation plus ${}^{n+p}C_p$, i.e. $${}^nC_0+{}^{n+1}C_1+{}^{n+2}C_2+\dots+{}^{n+p-1}C_{p-1}+{}^{n+p}C_p = {}^{n+p}C_{p-1} + {}^{n+p}C_{p}$$ By the addition property of combinations on the right hand side above, we get $${}^nC_0+{}^{n+1}C_1+{}^{n+2}C_2+\dots+{}^{n+p-1}C_{p-1}+{}^{n+p}C_p ={}^{n+p+1}C_p$$
Ahmad Bazzi
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