$$^{n-1}C_2+^{n-1}C_3+^nC_4+^{n+1}C_5+^{n+2}C_6+^{20}C_6$$ I Can't figure out how to solve it. I tried writing out the combinations' formula and then solving it like a normal equation, but I ended up with like a 6th degree polynomial.
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6Please put the full question and your attempt in the body of the question, not in an external link. – Jair Taylor May 13 '24 at 14:08
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1There is probably a misprint : intuitively, the first coefficient should be with $n-2$ not $n-1$ for symmetry reasons giving a final not-to-complicated equation. – Jean Marie May 13 '24 at 14:31
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Remind yourself of the hockeystick identity and compare the given expression to the hockey stick identity. – JMoravitz May 13 '24 at 14:32
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1@JeanMarie The expression had we continued to $\binom{n-2}{2}$ instead of $\binom{n-1}{2}$ is indeed useful to consider though it would be most useful to us had it continued through to the bottom number reaching zero, but here we notice that $\binom{n-4}{0}+\binom{n-3}{1}+\binom{n-2}{2} = \binom{n-1}{2}$ and so it works out as we expect. – JMoravitz May 13 '24 at 14:36
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@JMoravitz You are right. – Jean Marie May 13 '24 at 14:37