TITLE: Generating function of $\binom{3n}{n}$
QUESTION [8 upvotes]: Wolfram alpha tells me the ordinary generating function of the sequence $\{\binom{3n}{n}\}$ is given by
$$\sum_{n} \binom{3n}{n} x^n = \frac{2\cos[\frac{1}{3}\sin^{-1}(\frac{3\sqrt{3}\sqrt{x}}{2})]}{\sqrt{4-27x}}$$
How do I prove this?

REPLY [2 votes]: Too long for a comment: By investigating series of the form $S_a(x)=\displaystyle\sum_{n=0}^\infty{an\choose n}~x^n$, we notice that — in general — we get a $($ generalized $)$ hypergeometric function of argument $\dfrac{a^a}{(a-1)^{a-1}}~x$, which then naturally leads us to suspect that we should rather be inspecting the values of the same expression for $x(t)=\dfrac{(a-1)^{a-1}}{a^a}~t$. Taking into consideration the fact that $\displaystyle\lim_{u\to0}u^u=1$, we have 

$$\begin{align}
S_{-1}\Big(x(t)\Big)~&=~\frac1{\sqrt{1-t}}
\\\\
S_{-1/2}\Big(x(t)\Big)~&=~\frac{\cos\bigg(\dfrac{\arcsin t}3\bigg)+\dfrac1{\sqrt3}~\sin\bigg(\dfrac{\arcsin t}3\bigg)}{\sqrt{1-t^2}}
\\\\
S_0\Big(x(t)\Big)~&=~0
\\\\
S_{1/2}\Big(x(t)\Big)~&=~1-i~\frac{t}{\sqrt{1-t^2}}
\\\\
S_1\Big(x(t)\Big)~&=~\frac1{1-t}
\\\\
S_{3/2}\Big(x(t)\Big)~&=~\frac{\cos\bigg(\dfrac{\arcsin t}3\bigg)+\sqrt3~\sin\bigg(\dfrac{\arcsin t}3\bigg)}{\sqrt{1-t^2}}
\\\\
S_2\Big(x(t)\Big)~&=~\frac1{\sqrt{1-t}}
\\\\
S_3\Big(x(t)\Big)~&=~\frac{\cos\bigg(\dfrac{\arcsin\sqrt t}3\bigg)}{\sqrt{1-t}}
\end{align}$$