<s>[INST] Evaluate$\sum_{n=0}^\infty\binom{3n}{n}x^n$ The question is:

Evaluate $$\sum_{n=0}^\infty\binom{3n}{n}x^n$$

After applying a few numbers as $x$ in Wolfram Alpha, I guess that the answer is probably:
$$2\sqrt{\frac1{4-27x}}\cos\left( \frac13\sin^{-1}\frac{3\sqrt{3x}}{2} \right)$$
that I can never prove. (Interestingly the above becomes simply $2\cos\frac{\pi}9$ when $x=\frac19$.)
(*) To give you the background, the motivation that led me to this question is the Algebra problem #$10$ in the Harvard-MIT Math Test in Feb. 2008, that concludes to:
$$\sum_{n=0}^\infty\binom{2n}{n}x^n=\frac1{\sqrt{1-4x}}$$
And then I thought about what if it was $3n$ instead of $2n$.
 [/INST] This is too long for a comment.
Considering $$f_k=\sum_{n=0}^\infty\binom{kn}{n}x^n$$ where $k$ is a positive integer,we could notice nice patterns using the generalized hypergeometric function
$$f_4=\, _3F_2\left(\frac{1}{4},\frac{2}{4},\frac{3}{4};\frac{1}{3},\frac{2}{3};\frac{4^4
   }{3^3}x\right)$$
$$f_5=\,
   _4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{1}{4},\frac{2}{
   4},\frac{3}{4};\frac{5^5 }{4^4}x\right)$$
$$f_6=\,
   _5F_4\left(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};\frac{1}{
   5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{6^6 }{5^5}x\right)$$ As you already noticed, this only simplifies for $k=2$ and $k=3$.
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