Question

Use partial fractions to express as a power series: 3/(x^2-x-2)

i'm having trouble expressing as a power series. I get the partial fractions part.

Answer 1

So far I have gotten this via geometric series and partial fractions since A = 1 and B = -1. I just have no clue how to piece the two together.
\[\frac{1}{2} \times \sum_{n=0}^{\infty}(\frac{x}{2})^n-\sum_{n=0}^{\infty}(-x)^n\]

Answer 2

The form of power series looks like \(\displaystyle \sum c_n(x-a)^n\), right?
So...
\[
\begin{array}~
\dfrac{1}{2}\sum_{n=0}^\infty\left(\dfrac{x}{2}\right)^n-\sum_{n=0}^\infty(-x)^n &= \sum_{n=0}^\infty\dfrac{x^n}{2^{n+1}}-\sum_{n=0}^\infty(-x)^n\\~\\ &=\sum_{n=0}^\infty\dfrac{x^n}{2^{n+1}}-(-1)^nx^n\\~\\&=\sum_{n=0}^\infty x^n\left({\dfrac{1}{2^{n+1}}}-(-1)^n\right)\\~\\&=\boxed{\displaystyle \sum_{n=0}^\infty \left({\dfrac{1}{2^{n+1}}}+(-1)^{n+1}\right)x^n}
\end{array}
\]

That is good enough, right?

Answer 3

oh duhhhhhhhhhhhhh now i see how it works! thanks a lot i was stuck on this for about an hour lol.

Answer 4

Haha glad I helped.