Prove that $${2n \choose n} = \frac{1}{n+1} {2n \choose n} + \sum_{k=0}^{n-1}{2 \choose k}{2n-2k-1 \choose n-k}, \forall n \in \mathbb{N}$$
I am trying to prove this by using lattice path argument but can't come up with a solid solution. Any hel would be appreciated. Thank you
If we are talking about the same question, I think the full equation is
${2n \choose n} =\frac{1}{n + 1}{2n \choose n}+\sum_{k = 0}^{n - 1} \frac{1}{k + 1} {2k \choose k} { 2n - 2k - 1 \choose n - k}$
Apply the result of Catalan number to $\frac{1}{k + 1} {2k \choose k}$, the second part of RHS actually counts the number of lattice paths which immediately go above the diagonal after point $(k,k)$.