<s>[INST] If $y = \frac{2}{5}+\frac{1\cdot3}{2!} \left(\frac{2}{5}\right)^2+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^3+\cdots$, find $y^2+2y$ 
If $$y = \frac{2}{5}+\frac{1\cdot3}{2!} \left(\frac{2}{5}\right)^2+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^3+\cdots$$ what is $y^2+2y$?

Attempt:
We know that for negative and fractional indices,
$$(1+x)^n = 1 + nx + n(n-1)/2!\cdot x^2 + n(n-1)(n-2)/3!\cdot x^3 + \cdots$$
Rewriting the series in question, we get:
$$\frac{2}{5} \left(1 + \frac{1\cdot3}{2!}\cdot \frac{2}{5}+\frac{1\cdot3\cdot5}{3!} \left(\frac{2}{5}\right)^2+\cdots\right)$$
I know this looks like the binomial expansion above, but I have no idea how to proceed further.
 [/INST] You can calculate $y$ using a "well known" binomial series expansion for
$$(1-4x)^{-\frac 12} = \sum_{n=0}^{\infty}\binom{2n}nx^n$$
To see this rewrite the coefficients
$$\prod_{k=1}^n\frac{2k-1}{k} = \prod_{k=1}^n\frac{(2k-1)\cdot 2k}{k\cdot 2k}= \frac 1{2^n}\binom{2n}n$$
So in your case the series becomes
$$y = \sum_{\color{blue}{n=1}}^{\infty}\frac 1{2^n}\binom{2n}n \frac{2^n}{5^n} = \sum_{\color{blue}{n=1}}^{\infty} \binom{2n}n\frac 1{5^n}$$ $$ = \left.\frac 1{\sqrt{1-4x}}-1\right|_{x = \frac 15} = \sqrt 5 - 1$$
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