How to prove $\sum_{n=0}^\infty \left(\frac{(2n)!}{(n!)^2}\right)^3\cdot \frac{42n+5}{2^{12n+4}}=\frac1\pi$? In an article about $\pi$ in a popular science magazine I found this equation printed in light grey in the background of the main body of the article:
$$
\color{black}{
\sum_{n=0}^\infty \left(\frac{(2n)!}{(n!)^2}\right)^3\cdot \frac{42n+5}{2^{12n+4}}=\frac1\pi
}
$$
It's true, I checked it at Wolfram, who gives a even more cryptic answer at first glance, but finally confirms the result.
The appearance of $42$ makes me confident that there is someone out there in this universe, who can help to prove that?

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$\large\tt Hint:$
$\ds{%
{1 \over \pi}
=
\sum_{n=0}^{\infty}\bracks{\pars{2n}! \over \pars{n!}^{2}}^{3}
{42n + 5 \over 2^{12n + 4}}
=
{21 \over 8}\sum_{n=0}^{\infty}{2n \choose n}^{3}n\pars{2^{-12}}^{n}
+
{5 \over 16}\sum_{n=0}^{\infty}{2n \choose n}^{3}\pars{2^{-12}}^{n}\,,
\qquad{\large ?}}$
Let's consider the function
$\ds{{\cal F}\pars{x} \equiv \sum_{n=0}^{\infty}{2n \choose n}^{3}x^{n}}$ and we
have to evaluate
$\ds{\braces{\bracks{{21 \over 8}\,x\,\partiald{}{x}
+ {5 \over 16}}{\cal F}\pars{x}}_{x = 2^{-12}}}$ $\ds{\pars{~\mbox{this expression returns the value}\ {1 \over \pi}~}}$:

\begin{align}
{\cal F}\pars{x} &\equiv \sum_{n=0}^{\infty}x^{n}\int_{\verts{z_{1}} = 1}
{\dd z_{1} \over 2\pi\ic}\,{\pars{1 + z_{1}}^{2n} \over z_{1}^{n + 1}}
\int_{\verts{z_{2}} = 1}
{\dd z_{1} \over 2\pi\ic}\,{\pars{1 + z_{2}}^{2n} \over z_{2}^{n + 1}}\int_{\verts{z_{1}} = 1}
{\dd z_{1} \over 2\pi\ic}\,{\pars{1 + z_{3}}^{2n} \over z_{3}^{n + 1}}
\\[3mm]&=
\prod_{i = 1}^{3}\pars{\int_{\verts{z_{i}} = 1}
{\dd z_{i} \over 2\pi\ic}\,{1 \over z_{i}}}\sum_{n = 0}^{\infty}\bracks{%
x\pars{1 + z_{1}}^{2}\pars{1 + z_{2}}^{2}\pars{1 + z_{3}}^{2}
\over
z_{1}z_{2}z_{3}}^{n}
\\[3mm]&=
\prod_{i = 1}^{3}\int_{\verts{z_{i}} = 1}
{\dd z_{i} \over 2\pi\ic}\,
{1 \over
z_{1}z_{2}z_{3} - x\pars{1 + z_{1}}^{2}\pars{1 + z_{2}}^{2}\pars{1 + z_{3}}^{2}}
\end{align}

