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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!

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1 Answer 1

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The coefficient of $x^n$ in $$e^{2x}=\sum_{r=0}^\infty\dfrac{x^r2^r}{r!}=?$$

Now using Binomial series $$(1+x)^{-2}=1+\sum_{r=1}^\infty\dfrac{x^r(-2)(-3)\cdots(-r)(-r-1)}{r!}=1+\sum_{r=1}^\infty(-1)^r(r+1)x^r$$

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  • $\begingroup$ Genius! Thank you very much. I'll expand the Taylor Series and combine them to get an answer. $\endgroup$
    – user102938
    Commented May 12, 2020 at 9:57

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