Maclaurin series for $f(x)=\frac{1}{1+x+x^2} $

Question

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.

The 1 corresponds to terms $1, x^3, x^6,...$ and the $-x$ corresponds to terms $-x, -x^4, -x^7,...$ — Zarrax, Jan 28 '14 at 16:13
Possible duplicate of Find the power series of $f(x)=\frac{1}{x^2+x+1}$ — , May 20 '18 at 09:22

score 3 · Answer 1 · answered Jan 28 '14 at 15:39

3

HINT:

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

answered Jan 28 '14 at 15:39

lab bhattacharjee

274,582

score 1 · Answer 2 · answered Jan 28 '14 at 16:41

1

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

answered Jan 28 '14 at 16:41

Yiorgos S. Smyrlis

83,933

How does this answer differ from lab's, provided an hour ago? – Ron Gordon Jan 28 '14 at 16:42

Maclaurin series for $f(x)=\frac{1}{1+x+x^2} $

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