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.
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 $ .