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?

This is a famous identity of Ramanujan in "Modular equations and approximations of $\pi$".
There is a proof by the Borweins in "Pi and the AGM" (no preview) p. $177$ to $188$ (this proof and others are rather long!).
UPDATE:
"Ramanujan’s Series for 1/π: A Survey" provides the history of the subject with all the technical details.
The brothers Borwein proposed a derivation in $1987$ in "Ramanujan's rational and algebraic series for $\dfrac 1{\pi}$".
Guillera proposed different "Kind of proofs of Ramanujan-like series" in $2012$.
A proof 'by computer' using the WZ algorithm may be found in the paper of Ekhad and Zeilberger "A WZ proof of Ramanujan's formula for $\pi$".
Aycock proposes to compute many similar series using hypergeometric identities like (page $6$ and $28$) :
$$_3F_2\left(\frac12,\frac12,\frac12,1,1,x\right)=(1-x)^{-1/2}\;_3F_2\left(\frac14,\frac34,\frac12,1,1,-\frac{4x}{(1-x)^2}\right)$$
