Catalan number's combinatoric proof

Asked May 16 '23 at 16:56

Active May 16 '23 at 18:23

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prove a combinatoric proof the identity: $ \sum_{k=0}^{n} $$\frac{1}{k+1}\binom{2k}{k}\binom{2n-2k}{n-k} = \binom{2n+1}{n} $

I tried using Catalan numbers: and one of the formulas for them:

$C_{n} = $$\frac{1}{n+1}\binom{2n}{n}$ but it lead nowhere.

I tried opening the binom and maybe simplifying but it wasn't any good

edited May 16 '23 at 18:23

asked May 16 '23 at 16:56

yahel amity

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If I were a betting man, I would bet that this is a duplicate question. – Adam Rubinson May 16 '23 at 17:17
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If the question is "Any help?" - then this is not really a mathematical question. Please provide a good context, and really show the own progress. See also https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question to know how to provide a good context. – dan_fulea May 16 '23 at 18:03
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Does this answer your question? Proving this identity $\sum_k\frac{1}{k}\binom{2k-2}{k-1}\binom{2n-2k+1}{n-k}=\binom{2n}{n-1}$ using lattice paths – user May 17 '23 at 21:06

0 Answers0