<s>[INST] 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
 [/INST] Alternative approach: for $x$ near zero we have that $|x(3-2x)|<1$ and 
$$\frac{1}{1-3x+2x^2}=\frac{1}{1-x(3-2x)}=\sum_{k=0}^{\infty}(x(3-2x))^k\\=1+(x(3-2x))+(x(3-2x))^2+(x(3-2x))^3+o((x(3-2x))^3)\\
=1+(3x-2x^2)+(9x^2-12x^3+o(x^3))+(27x^3+o(x^3))+o(x^3)\\
=1+3x+7x^2+15x^3+o(x^3).$$
</s>