# For the power series $\sum_{n=0}^{\infty}c_nx^n=e^{2x}-\frac{3}{(1+x)^2}$, with $|x|<1$, find $c_n$

Problem: For the power series $$\sum_{n=0}^{\infty}c_nx^n=e^{2x}-\frac{3}{(1+x)^2}$$, with $$|x|<1$$, find $$c_n$$.

I made some beginnings, but unlike geometric series I don't think there is a closed form expression for the sum of a convergent power series. I tried doing some integration/differentiation but that seemed to get me nowhere. Am I right in saying that if this converges to $$e^{2x}-\frac{3}{(1+x)^2}$$, it must be that $$\sum_{n=0}^{\infty}c_nx^n$$ can be represented as a geometric series $$\sum_{n=0}^{\infty}ar^n$$ with $$r<1$$? This would mean that $$\frac{a}{1-r}=e^{2x}-\frac{3}{(1+x)^2}$$, which two variables but only one equation. Can I just pick $$a=1$$ and solve? If I do, I get a promising but extremely messy result. What's a better way to go about this? Any hints? Help would be much appreciated!

The coefficient of $$x^n$$ in $$e^{2x}=\sum_{r=0}^\infty\dfrac{x^r2^r}{r!}=?$$
Now using Binomial series $$(1+x)^{-2}=1+\sum_{r=1}^\infty\dfrac{x^r(-2)(-3)\cdots(-r)(-r-1)}{r!}=1+\sum_{r=1}^\infty(-1)^r(r+1)x^r$$