Context: Reading this paper: https://arxiv.org/abs/1309.1140. (On proving some of Ramanujan's formulas for $1/\pi$ with an elementary method). We find on page 26 that the author claims: $$\sum_{n=0}^{\infty}{\left({1\over 6}\right)_{n}\left({1\over 2}\right)_{n}\left({5\over 6}\right)_{n}\over(1)_{n}^3}(256n+20)\left({2\over 11}\right)^{3n} =\frac{11\sqrt{33}}{\pi},$$
which is not true which has a typo. (Note: As pointed out by Paramanand Singh, it should be $252n+20 = 4(63n+5).$ See corrected formula below.)
Following the translation method, we find:
$$\sum_{n=0}^{\infty}\frac{(6n)!}{n!^3(3n)!}\left(\frac{1}{66^3} \right)^n=\frac{\sqrt{33}\Gamma(1/4)^4}{32\pi^3},\tag{1}$$
and
$$\sum_{n=0}^{\infty}\frac{(6n)!n}{n!^3(3n)!}\left(\frac{1}{66^3} \right)^n=\frac{11\sqrt{33}}{252\pi}-\frac{5\sqrt{33}\Gamma(1/4)^4}{2016\pi^{3}}.\tag{2}$$
So we deduce the Ramanujan's series:
$$\sum_{n=0}^{\infty}\frac{(6n)!(63n+5)}{n!^3(3n)!}\left(\frac{1}{66^3} \right)^n=\frac{11\sqrt{33}}{4\pi},\tag{3}$$
using only hypergeometric transformations.
Question 1:
Why does Mathematica unable to deal with $(1)$, $(2)$ and $(3)$ when it is able to deal with more complicated Ramanujan's series?
Question 2:
Could you find a solution to this? Modular approach is also welcome.