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The Catalan numbers form the sequence of numbers starting 1,1,2,5,14,42,... with explicit formula $\frac{1}{n+1}\binom{2n}{n}$. It counts many combinatorial objects like planar binary trees, triangulations, noncrossing partitions, Dyck paths, etc. See https://oeis.org/A000108

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Why is this alternating sum involving Catalan numbers $\sum_{i=0}^{\lfloor t/2 \rfloor} (-1)...

I need the result that for all $t$, $$\sum_{i=0}^{\lfloor t/2 \rfloor} (-1)^{i+1} \binom{t-i}{i} C_{t-i-1} = 0,$$ where $C_{t-i-1}$ is the $(t-i-1)$-th Catalan number. I've checked for $t$ up to 100 …
Aaron Li's user avatar