TITLE: Looking for a "cute" justification for a Catalan-type generating function
QUESTION [8 upvotes]: The Catalan numbers $C_n=\frac1{n+1}\binom{2n}n$ have the generating function
$$c(x)=\frac{1-\sqrt{1-4x}}{2x}.$$
Let $a\in\mathbb{R}^+$. It seems that the following holds true
$$\frac{c(x)^a}{\sqrt{1-4x}}=\sum_{n=0}^{\infty}\binom{a+2n}nx^n.$$

QUESTION. Why?

REPLY [3 votes]: Let $C_a(x)=\frac{c(x)^a}{\sqrt{1-4x}}$ and $B_a(x) =\sum_{n=0}^{\infty}\binom{a+2n}nx^n.$
The identity 
$c(x)=1+xc(x)^2$ implies $C_{a+1}(x)= C_{a}(x)+x C_{a+2}(x).$
The recursion for the binomial coefficients implies
$B_{a+1}(x)= B_{a}(x)+x B_{a+2}(x)$.
If we show that  $B_a(x)=C_a(x)$ holds for $a=1$ then it holds for all positive integers. 
This follows from $B_1(x)=\frac{1}{2} \sum_{n=0}^{\infty}\binom{2+2n}{n+1}x^n=
\frac{1}{2x}(B_0(x)-1)=C_1(x).$