We know that the approximates by partial continued fractions of the golden ratios are the quotients of successive **Fibonacci numbers** $1,1,2,3,5,8,13,...$. Taking
$$
\phi=\frac{1+\sqrt{5}}2\approx \frac{8}5\implies \sqrt5\approx \frac{11}{5}
$$
we get a lower approximation for $\sqrt5$. To the remainder of the square root after factoring out the approximation we can apply the the Newton/binomial series for the inverse square root
$$
\sqrt5=\frac{11}{5}\sqrt{\frac{125}{121}}=\frac{11}{5}\left(1-4\frac1{5^3}\right)^{-1/2}
=\frac{11}5\sum_{n=0}^\infty\binom{2n}{n}\frac1{5^{3n}}
$$
which is the series you got.

Exploring the same method for one place further in the Fibonacci sequence
$$
\phi=\frac{1+\sqrt{5}}2\approx \frac{13}8\implies \sqrt5\approx \frac{9}{4}
$$
gives an upper approximation of $\sqrt5$ and thus an alternating series,
$$
\sqrt5=\frac{9}{4}\sqrt{\frac{80}{81}}=\frac{9}{4}\left(1+4\frac1{320}\right)^{-1/2}
=\frac{9}4\sum_{n=0}^\infty\binom{2n}{n}\frac{(-1)^n}{320^{n}}
$$

### Why that series?

The general **binomial series** reads as
$$(1+x)^α=\sum_{n=0}^\infty\binom{α}n.$$ For $α=-1/2$ the binomial coefficient can be transformed as
$$
\binom{-1/2}{n}=\frac{(-1/2)(-1/2-1)...(-1/2-n+1)}{n!}=(-1/2)^n\frac{(2n-1)(2n-3)...3\cdot 1}{n!}=(-1/4)^n\frac{(2n)!}{(n!)^2},
$$
resulting in the "simplified" formula
$$\frac1{\sqrt{1-4x}}=\sum_{n=0}^\infty\binom{2n}nx^n.$$