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What is the expansion for $\frac{1}{1+x+x^{2}}$? I know expansion for $$\begin{align}&(1+x)^{-1}=1-x+x^{2}-x^{3}+\dots\\ &(1+x)^{-2}=1+(-2)x+(-2)\frac{-3}{2!}+(-2)(-3)\frac{-4}{3!}+\dots\\ &(1+x^{2})^{-1}=1-x^{2}+x^{4}-x^{6}+\dots\end{align}.$$ But for $\frac{1}{1+x+x^{2}}$ I got problem. Can someone derive $\frac{1}{1+x+x^{2}}$ term expansion.

Robert Z
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Elizabeth
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1 Answers1

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Hint. Note that for $|x|<1$, $$\frac{1}{1+x+x^{2}}=\frac{1-x}{1-x^{3}}=(1-x)(1+x^3+x^6+x^9+\dots).$$

Robert Z
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  • thank u.Can u explain how u got $\frac{1}{1+x+x^{2}}=\frac{1-x}{1-x^{3}}$ – Elizabeth May 20 '18 at 09:38
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    @Elizabeth $a^3 - b^3 = (a - b)(a^2 + ab + b^3)$, so $1 - x^3 = (1 - x)(1 + x + x^2)$, as you can check by direct multiplication. – N. F. Taussig May 20 '18 at 09:59
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    @Elizabeth Just verify that $1 - x^3 = (1 - x)(1 + x + x^2)$. More generally $1 - x^n = (1 - x)(1 + x +\dots+ x^{n-1})$ – Robert Z May 20 '18 at 10:32