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

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...

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