Finding the Taylor series of $1\over 1+x-2x^2$ centered at $0$ My attempt:
I know that the Taylor series for $f(x) = {1\over 1-x}$ is $\sum_{k=0}^\infty x^k$. We can rewrite $1\over 1+x-2x^2$ as ${1 \over 1 - (-x + 2x^2)} = f(-x+2x^2) = \sum_{k=0}^\infty (-x+2x^2)^k$. However, I was told that this is wrong by my TA, but I'm not sure why. Can anyone explain why this is wrong? What will be the correct answer?

$$\frac{1}{1 + x - 2x^{2}} = \frac{1}{(1 - x)(1 + 2x)}$$
You can use partial fractions to split this up into two rational functions:
$$\frac{1}{(1 - x)(1 + 2x)} = \frac{A}{1 - x} + \frac{B}{1 + 2x}$$
Some work yields $A = 1/3, B = 2/3$. So:
$$\frac{1}{1 + x - 2x^{2}} = \frac{1/3}{1 - x} + \frac{2/3}{1 - (-2x)} = \frac{1}{3} \cdot \sum_{n = 0}^{\infty}x^n + \frac{2}{3} \cdot \sum_{n = 0}^{\infty}(-2x)^{n}$$
And this simplifies to:
$$\sum_{n = 0}^{\infty} \left(\frac{1 + 2(-2)^{n}}{3}\right)x^n$$
