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I am unable to solve this specific problem.

The only "notable series expansion" I can use (and know) is $\sum^{+\infty}_0 x^n =$$1\over(1-x)$

I tried several things but none worked.

Writing $x\over(1+x-2x^2)$ as $x * {1 \over(1-(-x+2x^2))}$$= x \sum(-x+2x^2)^n$ did not help. Differentiation and integration also seem to not lead anywhere.

If it helps, the answer should be $\sum {(1 - (-1)^n * 2^n) \over 3} * x ^n$

Another thing is could not, and really tried, finding an online series representation calculator.

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    $\begingroup$ First factorize the denominator, then break the fraction as a sum of 2 fractions, both of which could be expanded by using the notable series expansion you have mentioned. $\endgroup$ – xbh Apr 20 at 23:36
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    $\begingroup$ Well, that solves it. I spent over 40min without thinking about it... Thank you $\endgroup$ – MTLaurentys Apr 20 at 23:41
  • $\begingroup$ You are welcome. What I have said is exactly the same as mentioned by the answer below, i.e. do the partial fractions decomposition first. $\endgroup$ – xbh Apr 20 at 23:44
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Do partial fractions on it and you'll be fine

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$$\frac{x}{(1+x-2x^2)}=\frac{1}{3 (1 - x)} - \frac{1}{3 (1 + 2 x)}$$

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