Problem

Source: 2024 CTST P13

Tags: combinatorics



For a natural number $n$, let $$C_n=\frac{1}{n+1}\binom{2n}{n}=\frac{(2n)!}{n!(n+1)!}$$be the $n$-th Catalan number. Prove that for any natural number $m$, $$\sum_{i+j+k=m} C_{i+j}C_{j+k}C_{k+i}=\frac{3}{2m+3}C_{2m+1}.$$ Proposed by Bin Wang