By using a geometric series and a factorisation, compute the first three terms of this given Taylor expansion By using the geometric series
$$\frac{1}{1-x} = \sum_{n=0}^\infty x^n$$
and the factorisation
$$\frac{1}{1-3x+2x^2}= \left(\frac{1}{1-2x}\right) \left(\frac{1}{1-x}\right)$$
compute the first three terms of the Taylor expansion of
$$\frac{1}{1-3x+2x^2} \text{ around } x=0$$
My theory of how to do this question is input the summation to the factorisation equation and solve so the summation equals 
$$\sum_{n=0}^\infty x^n = \frac{1-2x}{1-3x+2x^2}$$
and solve by starting with $x=0$ and going to $x=2$.
But I feel this may be too simple to be true. Are my workings correct?
Thanks in advance
 A: Still another approach: division by increasing powers of $x$
Perform the division of the numerator by the denominator up to the prescribed order $3$, truncating the computations at order $3$. It goes this way:
$$\begin{array}{rrrrr}
&&\color{red}1&{}\color{red}{+3x}&{}\color{red}{+ 7x^2} &{}\color{red}{+15x^3}\\
1-3x+2x^2 &\Big(&1\\%
&&-1&{}+3x &{}-2x^2\\
\hline 
&&&3x&{}-2x^2\\
&&& -3x &{}+9x^2&{}-6x^3\\
\hline 
&&&&{}+7x^2 &{}-6x^3 \\
&&&&{}-7x^2 &{}+21x^3 \\
\hline
&&&&&15x^3\end{array}$$ 
This way, you can prove the general expansion is
$$\sum_{n\ge 0}(2^{n+1}-1)x^n.$$
A: That could work, but an easier way is the following. Note that since
$$
\frac{1}{1-x} = \sum_{n=0}^\infty x^n 
$$
it follows that
$$
\frac{1}{1-2x} = \sum_{n=0}^\infty (2x)^n = \sum_{n=0}^\infty 2^nx^n = 1 + 2x + 4x^2 +8x^3 + \cdots
$$
Now, since
$$
\frac{1}{1 - 3x + 2x^2} = \frac{1}{1 - 2x}\frac{1}{1-x}
$$
we can expand the right-hand side to be
$$
\frac{1}{1 - 2x}\frac{1}{1-x} = (1 + 2x + 4x^2 + \cdots)(1 + x + x^2 + x^3 + \cdots)
$$
from which point we can collect terms appropriately to get the answer you're looking for.
The point is, you can multiply  convergent (or even non-convergent, if you feel like it!) series to obtain the series for the product. Since you know the first two series for this one...
A: Alternative approach: for $x$ near zero we have that $|x(3-2x)|<1$ and 
$$\frac{1}{1-3x+2x^2}=\frac{1}{1-x(3-2x)}=\sum_{k=0}^{\infty}(x(3-2x))^k\\=1+(x(3-2x))+(x(3-2x))^2+(x(3-2x))^3+o((x(3-2x))^3)\\
=1+(3x-2x^2)+(9x^2-12x^3+o(x^3))+(27x^3+o(x^3))+o(x^3)\\
=1+3x+7x^2+15x^3+o(x^3).$$
