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What is the Maclaurin expansion of $f(x)=\dfrac{1}{1+x+x^2} $? Thank you!

Edit:

By multiplying both terms with $ (1-x) $ I got to $\dfrac{1}{1-x^3}-\dfrac{x}{1-x^3}$. Is it correct to transform this to $ \sum_{i=0}^n x^{3n} (1-x) $? I somehow have the idea that there must be only one $x$ term in a Taylor series.

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  • $\begingroup$ The 1 corresponds to terms $1, x^3, x^6,...$ and the $-x$ corresponds to terms $-x, -x^4, -x^7,...$ $\endgroup$
    – Zarrax
    Jan 28, 2014 at 16:13
  • $\begingroup$ Possible duplicate of Find the power series of $f(x)=\frac{1}{x^2+x+1}$ $\endgroup$
    – user99914
    May 20, 2018 at 9:22

2 Answers 2

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HINT:

$$(1+x+x^2)(1-x)=1-x^3$$

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Hint. $$ \frac{1-x}{1-x^3}=(1-x)\sum_{n=0}^\infty x^{3n}. $$

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  • $\begingroup$ How does this answer differ from lab's, provided an hour ago? $\endgroup$
    – Ron Gordon
    Jan 28, 2014 at 16:42

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