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?
 A: $$\frac{1}{1 + x - 2x^{2}} = \frac{1}{(1 - x)(1 + 2x)}$$
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:
$$\sum_{n = 0}^{\infty} \left(\frac{1 + 2(-2)^{n}}{3}\right)x^n$$
A: To expand my comment from above: the Taylor series of a function $f$ is a series of the form
$$\sum_{k = 0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n$$
where $a$ is a real (complex) number, and $f^{(n)}{(a)}$ are also real (complex) numbers. The series $\sum_k (-x + 2x^2)^k$ does not meet this criterion.

Notice that
$$\frac{1}{1 + x - 2x^2} = \frac{1}{(2x + 1)(1 - x)} = \frac{A}{1 + 2x} + \frac{B}{1 - x}$$
for numbers $A, B$ that you can determine. Now you have two geometric series.
A: See the comment for why what you have written is not a power series (generally, we only want to see powers of $(x-a)$ for $a$ the center).
To be right, try 
$$
\frac{-1}{2x^2-x-1}=-\frac{1}{(2x+1)(x-1)}
$$
and then partial fractions to find functions you know how to write as power series centered at $0$.
