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?

Ah, 42 and The Hitchhiker's Guide to the Galaxy. Would you like to know how this is connected to the 24-dimensional Leech lattice? :)
Given the Ramanujan-type formula, 
$$\sum_{n=0}^\infty \left(\frac{(2n)!}{(n!)^2}\right)^3\cdot \frac{An+B}{(C)^{n+1/2}}=\frac1\pi$$
there are relatively simple expressions for $A,C$. Define,
$$A(\tau) = \sqrt{d\big(C(\tau)-64\big)}$$
and the 24th power of the Weber modular function as,
$$C(\tau) = \mathfrak{f}^{24}(2\tau)=\left(\frac{\eta^2(2\tau)}{\eta(\tau)\eta(4\tau)}\right)^{24}$$
with the Dedekind eta function,
$$\eta(\tau) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})$$ 
The connection between the appearance of 24 in $\eta(\tau)$ and the Leech lattice is quite well-known.
Example: Let $\tau = \frac{1}{2}\sqrt{-d},\; d = 7$, then,
$$A(\tau) =4\cdot42$$
$$C(\tau) = 2^{12}$$
$$e^{\pi\sqrt{7}} \approx 2^{12} - 24.06\dots$$
$$\frac{4(42n+B')}{(2^{12})^{n+1/2}}$$
so,
$$\sum_{n=0}^\infty \left(\frac{(2n)!}{(n!)^2}\right)^3\cdot \frac{42n+B'}{2^{12n+4}}=\frac1\pi$$
hence why, using a 24th power of an eta quotient, the number 42 appears.
