Looking for a "cute" justification for a Catalan-type generating function 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?

 A: Combining comments of @esg and myself, we have
$$\frac{c(x)^a}{\sqrt{1-4x}} = c(x)^a(xc(x))' = \frac{1}{(a+1)x^a}((xc(x))^{a+1})'$$
and thus
$$[x^n]\ \frac{c(x)^a}{\sqrt{1-4x}} = \frac{1}{a+1}[x^{n+a}]\ ((xc(x))^{a+1})'=\frac{n+a+1}{a+1} [x^n]\ c(x)^{a+1} $$
$$= \frac{n+a+1}{a+1}\frac{a+1}{a+1+2n}\binom{a+1+2n}{n}=\binom{a+2n}{n}.$$
A: 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 $B_1(x)=\frac{1}{2} \sum_{n=0}^{\infty}\binom{2+2n}{n+1}x^n=
\frac{1}{2x}(B_0(x)-1)=C_1(x).$
A: Here's a way to do it:
Recall that $C_n$ counts the number of lattice paths from $(0,0)$ to $(2n,0)$ taking only steps of the form $(1,\pm 1)$ that never goes below the $x$-axis; call this a Dyck path.  Further, $$\frac{1}{\sqrt{1 - 4x}} = \sum \binom{2n}{n}x^k$$ which counts the total number of paths from $(0,0)$ to $(2n,0)$; call this a bridge.  Also, $\binom{a+2n}{n}$ is the number of lattice paths (with the same step set) from $(0,0)$ to $(2n+a,a)$, since we have $a + 2n$ steps total with $n$ down steps (and thus $a + n$ up steps); call this an upward path.  
Every upward path can be decomposed into:  


*

*A bridge (up to the last time it hits $0$).

*A single up step

*A dyck path (up until the last time it hits $1$).

*another single step 

*a dyck path
and so on.
This provides a bijection from a single bridge with an $a$-vector of Dyck paths.  Since the generating function for a single bridge with $a$-vector of Dyck paths is exactly the left-hand-side of your equality, it must equal the right-hand side.  
