<s>[INST] Maclaurin series for $f(x) = \frac{1}{1+x+x^2}$ 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!
 [/INST] 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?
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