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

Question

I need to get the Maclaurin series and its radius of the convergence for $f(x) = \dfrac{1}{1+x+x^2}$. I tried to do it manually, with getting the derivatives, but I gave up after some time, because I thought there had to be a better way to solve this. Could anyone help? Thanks in advance!

Answer 1

Note that, if $\lvert x\rvert<1$,\begin{align}\frac1{1+x+x^2}&=\frac{1-x}{(1+x+x^2)(1-x)}\\&=\frac{1-x}{1-x^3}\\&=(1-x)\times(1+x^3+x^6+x^9+\cdots)\\&=1-x+x^3-x^4+x^6-x^7+x^9-x^{10}+\cdots\end{align}

Answer 2

Alternatively: $$\begin{align}\frac{1}{1+x+x^2}&=1-(x+x^2)+(x+x^2)^2-(x+x^2)^3+(x+x^2)^4-(x+x^2)^5+\cdots\\ &=1-x+x^2-x^3+x^4-x^5+x^6-x^7+\cdots\\ &\qquad -x^2+2x^3-3x^4+4x^5-5x^6+\cdots\\ &\qquad \ \ \ \ \ x^4-3x^5+6x^6-10x^7+\cdots\\ &\qquad -x^6+4x^7-10x^8+\cdots\\ &\qquad \ \ \ \ \ x^8-5x^9+\cdots\\ &\qquad -x^{10}+\cdots=\\ &=(1-x+0\cdot x^2)+(x^3-x^4+0\cdot x^5)+(x^6-x^7+0\cdot x^8)+\cdots.\end{align}$$ Do you see Pascal's triangle?

And also the pattern of coefficients $(1,-1,0)$ inside brackets at the end?