A question on generating function

Question

How to find the generating function of $\binom{2n}{n}$?

Answer 1

Let $F(z) = \sum\limits_{n \geq 0} \binom{2n}{n}z^n$. Then we can use what Thomas wrote in the comments: \begin{align*} F(z)^2 = F(z)\cdot F(z) &= \left( \sum\limits_{n \geq 0} \binom{2n}{n}z^n\right) \left( \sum\limits_{n \geq 0} \binom{2n}{n}z^n \right)\\ &= \sum\limits_{n \geq 0} \left(\sum\limits_{k = 0}^n \binom{2k}{k}\binom{2n - 2k}{n - k} \right) z^n \\ &= \sum\limits_{n \geq 0} 4^n z^n \\ &= \frac{1}{1 - 4z}. \end{align*}

This implies that $F(z) = \frac{1}{\sqrt{1 - 4z}}$.