TITLE: Find the Nth derivative using Maclaurin series
QUESTION [8 upvotes]: The prompt is to find the 8th derivative of the function f(x) defined as,
$$f(x) = \frac{3}{1+x-2x^2}$$
$$f^{(8)}(0) = ?$$ 
To find the maclaurin series, I proceeded by finding the derivatives of the function at 0 as follows,
$$f^{(1)}(x) = \frac{ -3 + 12x }{ (1 + x -2x^2)^2 } $$
$$f^{(2)}(x) = \frac{18(4x^2-2x+1)}{(-2x^2+x+1)^3} $$
such that, 
$$f(0) = 3$$
$$f^{(1)}(0) = -3$$
$$f^{(2)}(0) = 18$$
$$f^{(3)}(0) = -90$$
This makes the maclaurin series, 
$$f(x) = 3 - 3x + \frac{18x^2}{3!} - \frac{90x^3}{4!} + ...$$
I understand from the series, we have to have $(-1)^n$ since the negative sign is alternating, also in the denominator we have n! 
$$\sum_{n=0}^{\infty} \frac{(-1)^n(x)^n}{n!}$$ but it is incomplete, since I don't see the pattern, I would appreciate I someone can help with completing the series hence finding the 8th derivative.

REPLY [5 votes]: Just another way.
For the most general case of finding $f^{(n)}(0) = ?$, you received nice answers. Since I am lazy and since $n=8$ is quite small, I used long division to get 
$$\frac{3}{1+x-2x^2}=3-3 x+9 x^2-15 x^3+33 x^4-63 x^5+129 x^6-255 x^7+513 x^8+\cdots$$ Then $$f^{(8)}(0) =513\times 8!$$