How to prove this Catalan number identity

Question

Catalan number is $\displaystyle C_n= \frac{1}{n+1}\binom{2n}{n}$. How to prove that $$C_{2n-1} = \sum_{k=0}^{n-1}\left(\binom{2n-1}{n-k-1}-\binom{2n-1}{n-k-2}\right)^2$$ for $n\geq 1$. Thank you.

Answer 1

It seems that we can proceed as follows.

Put $$\Sigma=\sum_{k=0}^{n-1}\left(\binom{2n-1}{n-k-1}-\binom{2n-1}{n-k-2}\right)^2$$ and $$\Sigma’=\sum_{k=0}^{n-1}\left(\binom{2n-1}{n-k-1}+\binom{2n-1}{n-k-2}\right)^2.$$ Then $$\Sigma+\Sigma’=2\sum_{k=0}^{n-1}\binom{2n-1}{n-k-1}^2+\binom{2n-1}{n-k-2}^2=$$ $$2\sum_{k=0}^{2n-1}\binom{2n-1}{k}^2-2\binom{2n-1}{n}^2=$$ $$2\left(\binom{4n-2}{2n-1}-\binom{2n-1}{n}^2\right).$$

But $$\Sigma’=\sum_{k=0}^{n-1}\binom{2n}{n-k-1}^2=\frac 12\left(\sum_{k=0}^{2n}\binom{2n}{k}^2-\binom{2n}{n}^2 \right)= \frac 12\left(\binom{4n}{2n}-\binom{2n}{n}^2\right).$$

Substacting the formula for $\Sigma’$ from the formula for $\Sigma+\Sigma'$, we obtain exactly $\frac 1{2n}\binom{4n-2}{2n-1}$.