TITLE: Evaluate$\sum_{n=0}^\infty\binom{3n}{n}x^n$
QUESTION [9 upvotes]: The question is:

Evaluate $$\sum_{n=0}^\infty\binom{3n}{n}x^n$$

After applying a few numbers as $x$ in Wolfram Alpha, I guess that the answer is probably:
$$2\sqrt{\frac1{4-27x}}\cos\left( \frac13\sin^{-1}\frac{3\sqrt{3x}}{2} \right)$$
that I can never prove. (Interestingly the above becomes simply $2\cos\frac{\pi}9$ when $x=\frac19$.)
(*) To give you the background, the motivation that led me to this question is the Algebra problem #$10$ in the Harvard-MIT Math Test in Feb. 2008, that concludes to:
$$\sum_{n=0}^\infty\binom{2n}{n}x^n=\frac1{\sqrt{1-4x}}$$
And then I thought about what if it was $3n$ instead of $2n$.

REPLY [7 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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Lets $\ds{\,\mrm{f}\pars{x} = \sum_{n = 0}^{\infty}{3n \choose n}x^{n}.
\qquad\mrm{f}\pars{0} = 1}$.

Then,
\begin{align}
\mrm{f}'\pars{x} & =
\sum_{n = 1}^{\infty}{3n \choose n}nx^{n - 1} =
\sum_{n = 0}^{\infty}{3n + 3 \choose n + 1}\pars{n + 1}x^{n} =
\sum_{n = 0}^{\infty}{\pars{3n + 3}! \over n!\pars{2n + 2}!}\,x^{n}
\\[5mm] & =
\sum_{n = 0}^{\infty}{\pars{3n + 3}\pars{3n + 2}\pars{3n + 1}\pars{3n}! \over n!\pars{2n + 2}\pars{2n + 1}\pars{2n}!}\,x^{n} =
{3 \over 2}\sum_{n = 0}^{\infty}{\pars{3n + 2}\pars{3n + 1} \over 2n + 1}
{3n \choose n}\,x^{n}
\\[5mm] & =
{27 \over 8}\
\underbrace{\sum_{n = 0}^{\infty}{3n \choose n}\,x^{n}}_{\ds{\mrm{f}\pars{x}}}\ +\
{27 \over 4}\
\underbrace{\sum_{n = 0}^{\infty}{3n \choose n}\,n\,x^{n}}
_{\ds{x\,\mrm{f}'\pars{x}}}\ -\
{3 \over 8}\sum_{n = 0}^{\infty}{1 \over 2n + 1}{3n \choose n}\,x^{n}
\label{1}\tag{1}
\end{align}

where $\ds{\mrm{f}'\pars{0} = 3}$.

Note that
\begin{align}
\sum_{n = 0}^{\infty}{1 \over 2n + 1}{3n \choose n}\,x^{n} & =
\sum_{n = 0}^{\infty}{3n \choose n}\,x^{n}\int_{0}^{1}t^{2n}\,\dd t =
\int_{0}^{1}\sum_{n = 0}^{\infty}{3n \choose n}\,\pars{xt^{2}}^{n}\,\dd t =
\int_{0}^{1}\mrm{f}\pars{xt^{2}}\,\dd t
\\[5mm] & =
{1 \over 2}\,x^{-1/2}\int_{0}^{x}{\mrm{f}\pars{t} \over t^{1/2}}\,\dd t
\end{align}

Expression \eqref{1} is reduced to
\begin{align}
\pars{16x^{1/2} - 108x^{3/2}}\,\mrm{f}'\pars{x} & =
54x^{1/2}\,\mrm{f}\pars{x} -3\int_{0}^{x}{\mrm{f}\pars{t} \over t^{1/2}}\,\dd t
\end{align}
Moreover,
\begin{align}
&\pars{8x^{-1/2} - 162x^{1/2}}\,\mrm{f}'\pars{x} +
\pars{16x^{1/2} - 108x^{3/2}}\,\mrm{f}''\pars{x}
\\[5mm] = &\
27x^{-1/2}\,\mrm{f}\pars{x} + 54x^{1/2}\,\mrm{f}'\pars{x} -
3\,\mrm{f}\pars{x}x^{-1/2}
\end{align}
and
\begin{equation}\bbx{\ds{
\pars{16x - 108x^{2}}\,\mrm{f}''\pars{x} +
\pars{8 - 216x}\,\mrm{f}'\pars{x} - 24\,\mrm{f}\pars{x} = 0\,,\qquad
\left\{\begin{array}{rcl}
\ds{\mrm{f}\pars{0}} & \ds{=} & \ds{1}
\\[2mm]
\ds{\mrm{f'}\pars{0}} & \ds{=} & \ds{3}
\end{array}\right.}}\label{2}\tag{2}
\end{equation}
The solution of differential equation \eqref{2} is given by:
$$\bbox[#ffe,25px,border:1px dotted navy]{\ds{%
{\root{3} \over \root{4 - 27x}}\bracks{%
\cos\pars{{1 \over 3}\,\mrm{arccsc}\pars{2 \over \root{4 - 27x}}} +
\sin\pars{{1 \over 3}\,\mrm{arccsc}\pars{2 \over \root{4 - 27x}}}}}}
$$
An equivalent expresion is
$$\bbox[#ffe,25px,border:1px dotted navy]{\ds{%
{\root{6} \over \root{4 - 27x}}
\cos\pars{{1 \over 3}\,\mrm{arccsc}\pars{%
2 \over \root{4 - 27x}} - {\pi \over 4}}}}
$$