<s>[INST] Evaluate a series with binomial coefficients 
If 
  $$y=\frac{3}{5}+\frac{1\cdot 3}{2!}(\frac{2}{5})^2+\frac{1\cdot 3\cdot 5}{3!}(\frac{2}{5})^3+\dots$$
  Then find the value 0f $y^2+2y$.

My approach is as follow
$$y-\frac{1}{5}=\frac{2}{5}+\frac{1\cdot 3}{2!}(\frac{2}{5})^2+\frac{1\cdot 3\cdot 5}{3!}(\frac{2}{5})^3+\dots$$
Let $x=\frac{2}{5}$, $Y=y-\frac{1}{5}$. Then
$$Y=x+\frac{1\cdot 3}{2!}(x)^2+\frac{1\cdot 3\cdot 5}{3!}(x)^3+\dots$$
I am not able to proceed from here.
 [/INST] Hint. Recall the definition of double factorial, 
$$1\cdot 3\cdots (2k-1)=\frac{k!}{2^k}\binom{2k}{k}.$$
Hence
$$Y=x+\frac{1\cdot 3}{2!}(x)^2+\frac{1\cdot 3 \cdot 5}{3!}(x)^3+\dots=
\sum_{k=1}^{\infty}\binom{2k}{k}(x/2)^k.$$
Now take a look at How to show that $1 \over \sqrt{1 - 4x} $ generate $\sum_{n=0}^\infty \binom{2n}{n}x^n $ .
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