The formula
$$\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=\frac{288\sqrt{5}}{5\pi}$$
(in older notation) appears as eq. 38 in Ramanujan's paper *Modular equations and approximations to $\pi$*; $(a)_n$ is the Pochhammer symbol.

But – the formula is like an exercise for the reader. Allegedly, it can be deduced from the theory of modular equations, but how exactly?

It should be somehow possible to prove the formula from Clausen's formulas and representations of $$25P(q^{50})-P(q^2)$$ where $$P(q)=1-24\sum_{k=1}^\infty \frac{kq^k}{1-q^k}$$ with $|q|\lt 1$. The hypergeometric representation of Ramanujan's formula is known to be $$41\,_3F_2\left(\frac14,\frac12,\frac34,1,1,-\frac{1}{25920}\right)-\frac{161}{69120}\, _3F_2\left(\frac54,\frac32,\frac74,2,2,-\frac{1}{25920}\right)=\frac{288\sqrt{5}}{5\pi}.$$

**What I tried**

Let $$K(x)=\int_0^{\pi/2}\dfrac{dt}{\sqrt{1-x\sin^2 t}}$$ and $$E(x)=\int_0^{\pi/2}\sqrt{1-x\sin^2 t}\,dt$$ be the elliptic integrals. Using the hypergeometric differential equation, the problem at hand shoud then be reducible to $$K\left(\dfrac{1}{2}-6\sqrt{-360+161\sqrt{5}}\right)\left(2E\left(\dfrac{1}{2}-6\sqrt{-360+161\sqrt{5}}\right)-\left(1+\sqrt{-840+376\sqrt{5}}\right)K\left(\dfrac{1}{2}-6\sqrt{-360+161\sqrt{5}}\right)\right)=\frac{\pi}{10}.$$

Now I recognized the argument of the elliptic integrals as a special value of the modular lambda function $\lambda$ (https://en.wikipedia.org/wiki/Modular_lambda_function): $$\lambda (5i)=\dfrac{1}{2}-6\sqrt{-360+161\sqrt{5}}.$$

So if someone can prove that $$K(\lambda (5i))=\dfrac{\sqrt{5}+2}{20}\dfrac{\Gamma (1/4)^2}{\sqrt{\pi}}$$ and $$E(\lambda (5i))=\dfrac{(-2+\sqrt{5})\pi^{3/2}}{\Gamma (1/4)^2}+\dfrac{\left(2+\sqrt{5}+2\sqrt{-10+6\sqrt{5}}\right)\Gamma (1/4)^2}{40\sqrt{\pi}},$$ (that's out of my reach), then Ramanujan's formula is proved.

**Edit**

I noticed that $K(\lambda (5i))$ can be easily proved to have the desired closed form by using division values of elliptic functions with parameter $-1$. But evaluating $E(\lambda (5i))$ seems hard...

This question was also asked on MSE: https://math.stackexchange.com/questions/4898590/how-did-ramanujan-find-sum-n-0-infty-1n-frac1-2-n1-4-n3-4-nn