How can I find the Maclaurin series of $\frac{1}{1+x+x^2}$?

Question

How can I find the Maclaurin series of $\frac{1}{1+x+x^2}$? At first, I found the Maclaurin series of $\frac{1}{1+x}$, which is $\sum_{n=0}^{\infty}(-1)^{n}x^{n}$ and simply replaced $x$ with $x^2 + x $, but the answer sheet says something different. It says that the answer is $\sum_{n=0}^{\infty}x^{3n}$. Which answer is right? The answer sheet? or me?

Answer 1

Hint: When $x\ne 1$, our function is $\frac{1-x}{1-x^3}$. Expand $\frac{1}{1-x^3}$ by using the series for $\frac{1}{1-t}$, and multiply by $1-x$.

Remark: The answer sheet, if quoted correctly, is not right. Your method works for getting the first few terms of the Taylor series, but becomes awkward for the general term. However, combined with the method of the hint, it can be used to prove a nice identity.