A Ramanujan infinite series $$ 1-5\left(\frac{1}{2}\right)^3+9\left(\frac{1 \cdot 3}{2 \cdot 4}\right)^3-13\left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right)^3+\cdots $$
I went on evaluating the above series and encountered that solving $\displaystyle \sum_{n\ge 0}\left(\binom{2n}{n}\right)^3x^n$ would suffice.
But how do we make a generating function for the third power of a central binomial coefficient using the fact $\displaystyle \sum_{n\ge 0}\binom{2n}{n}x^n=\frac{1}{\sqrt{1-4x}}$

Modifying @Claude Leibovici's answer a little.
$\frac{1}{\left(1-z\right)^{a}}=\,_{1}F_{0}\left(a;;z\right)={\displaystyle \sum_{n=0}^{\infty}\frac{a_{n}}{n!}z^{n}}$
and
$\frac{1}{\sqrt{1-4\cdot x}}={\displaystyle \sum_{n=0}^{\infty}\left(\begin{array}{c}
2n\\
n
\end{array}\right)x^{n}}$
Let $z=4x,x=\frac{z}{4}$
$\frac{1}{\left(1-4x\right)^{a}}=\,_{1}F_{0}\left(a;;4x\right)={\displaystyle \sum_{n=0}^{\infty}\frac{a_{n}}{n!}\,\left(4x\right)^{n}}$
$4^{n}\frac{\left(\frac{1}{2}\right)_{n}}{n!}=\left(\begin{array}{c}
2n\\
n
\end{array}\right)$
And we have the generating function for $\left(\begin{array}{c}
2n\\
n
\end{array}\right)^{k}\left(4^{k}x\right)^{n}$
i.e.
$\,_{k}F_{k-1}\left({\displaystyle \prod_{i=1}^{k}\left(\frac{1}{2}\right)_n};{\displaystyle \prod_{i=1}^{k-1}\left(1\right)_n};4^{k}z\right)={\displaystyle \sum_{n=0}^{\infty}}\left(\begin{array}{c}
2n\\
n
\end{array}\right)^{k}\left(4^{k}x\right)^{n}$
Where we “borrowed” the $n!=\left(1\right)_{n}$
