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Binomial expansion of $\frac{1}{1+x+x^2}$ up to the first three terms I am unsure where to start with this as i cannot put it into partial fractions, so don't really have an idea on where to start and pointers would be helpful.

H.Linkhorn
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    Hint: $\frac{1}{1+x+x^2} = \frac{1-x}{1-x^3}$. Can you find the Taylor expansion of $\frac{1}{1-x^3}$ ? – Jakobian Oct 14 '18 at 10:05
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    You can quite easily expand this rational function completely. Note that $$f(x):=\frac{1}{1+x+x^2}=\frac{1-x}{1-x^3}=(1-x),\sum_{k=0}^\infty,x^{3k}=1-x+x^3-x^4+x^6-x^7+x^9-x^{10}+\ldots,.$$ That is, for $k=0,1,2,\ldots$, the coefficient of $x^{3k}$ in $f(x)$ is $1$, the coefficient of $x^{3k+1}$ is $-1$, and the coefficient of $x^{3k+2}$ is $0$. – Batominovski Oct 14 '18 at 10:06
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    Oh, Jakobian already gave the hint, while I was writing my comment. – Batominovski Oct 14 '18 at 10:07
  • Possible Duplicate of https://math.stackexchange.com/questions/1990704 – Nosrati Oct 14 '18 at 10:20

2 Answers2

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$$\frac{1}{1+x+x^2}=\frac{1-x}{1-x^3} =(1-x)(1+x^3+x^6+...)$$

$$= 1-x+x^3 -x^4+...$$

Mohammad Riazi-Kermani
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As an alternative to the $\frac{1-x}{1-x^3}$ approach, you may know $\frac{1}{1-y}=1+y+y^2+y^3+y^4+\cdots$

so letting $y=-x-x^2$ you get $$\frac{1}{1+x+x^3} = 1 +(-x-x^2) +(-x-x^2)^2 + (-x-x^2)^3 + (-x-x^2)^4 + \cdots \\= 1 - x -x^2 + x^2 +2x^3 +x^4 - x^3 - 3x^4 - \cdots +x^4+\cdots \\ =1 - x +x^3 -x^4 +\cdots$$

Henry
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