Partial Fractions - combinatorics - trouble Having a very hard time with this question:
Q:  Use partial fractions to find the power series expansion of $$\frac{1+5x}{1-2x-3x^2}$$
 A: The partial fraction decomposition is 
$$-\frac{1}{1+x} - \frac{2}{-1+3x}$$
As the geometric series is 
$$\frac{1}{1-x}=\sum_{k=0}^\infty x^k $$ 
So we have
$$\frac{1+5x}{1-2x-3x^2}=-\frac{1}{1+x}+\frac{2}{1-3x}=-\sum_{k=0}^\infty 
(-1)^k x^k + 2\cdot \sum_{k=0}^\infty 3^k x^k=\sum_{k=0}^\infty (2 \cdot 3^k -(-1)^k) x^k  $$ 
As you see that the partial fraction is really helpful here  I will make it more explicit. we can use partial fraction decomposition if we know the zeroes of the denominator, when $x=-1$ we have 
$$1 - 2 (-1)-3(-1)^2=1+2-3=0$$ so $x=-1$ is a zero. the other zero is at $x=\frac{1}{3}$.
Now we have 
$$\frac{1+5x}{1-2x-3x^2}=\frac{1+5x}{(1+x)(1-3x)}$$
We try to write it as 
$$\frac{1+5x}{(1+x)(1-3x)}=\frac{A}{1+x} + \frac{B}{1-3x}$$
The way to achiefe $A$ and $B$ is the following: we know that 
$$\frac{A}{1+x} + \frac{B}{1-3x}= \frac{A(1-3x)+ B(1+x)}{1-2x-3x^2}$$
As we have 
$$\frac{A(1-3x)+ B(1+x)}{1-2x-3x^2}=\frac{x(B-3A)+1(A+B)}{1-2x-3x^2}$$
Ok now we compare the coeffients of $1+5x$ and $x(B-3A)+1(A+B)$ 
We have 
$$1=A+B$$ 
and 
$$5 = B-3A$$
I hope I made no mistake.
A: I just want to show how to apply Partial Fraction Decomposition
$$\frac{1+5x}{1-2x-3x^2}=\frac{1+5x}{(1-3x)(1+x)}=\frac A{1-3x}+\frac B{1+x}$$
$$\implies 1+5x=A(1+x)+B(1-3x)=x(A-3B)+A+B$$
Comparing the coefficients of the different powers of $x,A+B=1,A-3B=5$ 
$\implies B=-1,A=2$
Alternatively,
putting $x+1=0\implies x=-1,\quad 1+5\cdot(-1)=A\cdot0+B\cdot\{1-3(-1)\}\implies B=-1 $
putting $1-3x=0\implies x=\frac{1}{3},\quad 1+5\cdot\left(\frac13\right)=A\cdot\left(1+\frac13\right)+B\cdot0\implies A=2 $
