Help finding a proof of an identity $\sum_{k=1}^n(-1)^{n-k}{n \choose k}{n+k \choose k-1}=n$

Asked Sep 16 '22 at 11:29

Active Sep 16 '22 at 11:33

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Help finding a proof of an identity $\sum_{k=1}^n(-1)^{n-k}{n \choose k}{n+k \choose k-1}=n$. I have checked that the equality holds for n = 2,3,4,5,... etc. However I have no idea how to prove this identity for arbitrary n.

edited Sep 16 '22 at 11:33

asked Sep 16 '22 at 11:29

naoto

1

Does this answer your question? Prove that $\sum_{k=1}^a (-1)^{a-k} {a \choose k} {b+k \choose b+1} = {b \choose a-1}$ (It's the same identity since, taking $b = a$, $(a + k) - (a + 1) = k - 1$.) – Jakob Streipel Sep 16 '22 at 11:55
Perhaps induction over $n$ – weymar andres Sep 16 '22 at 11:55
prets: Yes, thank you very much! I haven't thought that we can use the Vandermode's identity for solving the problem. That's very cool!
– naoto Sep 16 '22 at 13:07

Help finding a proof of an identity $\sum_{k=1}^n(-1)^{n-k}{n \choose k}{n+k \choose k-1}=n$

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