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

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
$$\sum_{n\ge 0}(2^{n+1}-1)x^n.$$
