Coefficients of series given by generating function How to find the coefficients of this infinite series given by generating function.$$g(x)=\sum_{n=0}^{\infty}a_nx^n=\frac{1-11x}{1-(3x^2+10x)}$$
I try to expand like Fibonacci sequences using geometric series and binomial theorem but without any success.  
 A: I proposed the solution of a similar problem (
Generating Functions- Closed form of a sequence).
Write
$$
(1-3x^2-10x)\sum_{n=0}^{\infty}a_nx^n=1-11x
$$
and multiply. You get
$$
3a_{n-2}+10a_{n-1}-a_n=0
$$
with some initial conditions.
A: Hint: The denominator factors, though the roots are not very nice.  
Use the method of partial fractions to rewrite your expression.
You will get a sum of two simple terms, each of which has a nice expansion, derived from the infinite geometric series.
Remark: It is possible that you have the wrong denominator. For example, if it is really supposed to be a relative of $3x^2\pm x-10$, it would factor nicely. Often when one sets a problem, one tries to arrange things to make the arithmetic simple.
A: Use (1) the fact that $1-3x^2-10x=(1-ax)(1-bx)$ for some $a$ and $b$, (2) the fact that
$$
\frac{1-11x}{(1-ax)(1-bx)}=\frac{c}{1-ax}+\frac{d}{1-bx},
$$
for some $c$ and $d$, and (3) the fact that, for every $e$,
$$
\frac1{1-ex}=\sum_{n\geqslant0}e^nx^n.
$$
Then put all these pieces together to deduce that, for every $n\geqslant0$,
$$
a_n=c\cdot a^n+d\cdot b^n.
$$
Edit: You might want to note that $g(0)=1$ hence $a_0=1$, which yields $c+d=1$. 
Likewise, $1/(1-u)=1+u+o(u)$ when $u\to0$ hence $g(x)=(1-11x)(1+10x+o(x))=1-x+o(x)$. This shows that $g'(0)=-1$ thus $a_1=-1$ and $ca+(1-c)b=-1$, that is, $c=(1+b)/(b-a)$. Finally, $a$ and $b$ are the roots of the denominator $1-3x^2-10x$ and, for every $n\geqslant0$,
$$
a_n=\frac{(1+b)\cdot a^n-(1+a)\cdot b^n}{b-a}.
$$
