First five terms of power series

Question

I want to find first $5$ terms for power series of $\frac{1 - x + 2 x^2}{1 - 5 x^3 + 3 x^6}$.

$a_0 = \frac{1 - x + 2 x^2}{1 - 5 x^3 + 3 x^6} (0)$ = 1

$a_1 = \frac{\partial \frac{1 - x + 2 x^2}{1 - 5 x^3 + 3 x^6}}{\partial x}(0)$ = $-1$

$a_2 = \frac{1}{2!}\frac{\partial^2 \frac{1 - x + 2 x^2}{1 - 5 x^3 + 3 x^6}}{\partial x^2}(0)$ = $7$

$a_3 = \frac{1}{3!}\frac{\partial^3 \frac{1 - x + 2 x^2}{1 - 5 x^3 + 3 x^6}}{\partial x^3}(0)$ = $-5$

$a_4 = \frac{1}{4!}\frac{\partial^4 \frac{1 - x + 2 x^2}{1 - 5 x^3 + 3 x^6}}{\partial x^4}(0)$ = $35$

It is true?

Does more simple solution exists?

Answer 1

To get the series up to $x^5$:

$\begin{array}\\ \frac{1 - x + 2 x^2}{1 - 5 x^3 + 3 x^6} &=(1 - x + 2 x^2)\sum_{n=0}^{\infty} (5 x^3 - 3 x^6)^n\\ &=(1 - x + 2 x^2)(1+(5 x^3 - 3 x^6)+...)\\ &=(1 - x + 2 x^2)(1+5 x^3 )\\ &=(1 - x + 2 x^2)+(5x^3 - 5x^4 + 10 x^5)\\ &=1 - x + 2 x^2+5x^3 - 5x^4 + 10 x^5\\ \end{array} $