How to prove $\sum_{k=0}^{n-1}(-1)^k\binom{n+k}{2k+1}C_k=1$?

Asked Feb 14 '23 at 06:44

Active Feb 14 '23 at 06:44

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How to prove the following identity? $$ \sum_{k=0}^{n-1}(-1)^k\binom{n+k}{2k+1}C_k=1 $$ where $C_k$ is the $k$th Catalan number, $C_n=\frac{1}{k+1}\binom{2k}{k}$.

This question is similar to this but are different. For the problem in the link $m$ is in the second line of $\binom{s}{m-s}$, while for my problem $n$ is in the first line of $\binom{n+k}{2k+1}$. Therefore I tried the method in the link but it doesn't work.

Proofs, hints, or references are all welcome.

asked Feb 14 '23 at 06:44

bananana

1

Does this answer your question? Prove an equation about binomial coefficients - found using an Approach0 search. Note inserting the definition $C_k$ into the summation gives $\sum_{k=0}^{n-1}\frac{(-1)^k}{k+1}\binom{n+k}{2k+1}\binom{2k}{k}=1$, & your equation is a special case with $m=1$ since $\binom{n-1}{m-1}=1$. ... – John Omielan Feb 14 '23 at 07:25
(cont.) FYI, a quite similar problem in the results list is Calculate $\sum_{0 \le k } \binom{n+k}{2k} \binom{2k}{k} \frac{(-1)^k}{k+1}$. Last, but not least, welcome to Math SE. – John Omielan Feb 14 '23 at 07:30

How to prove $\sum_{k=0}^{n-1}(-1)^k\binom{n+k}{2k+1}C_k=1$?

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