Partial Fractions - combinatorics - trouble

Question

Having a very hard time with this question:

Q: Use partial fractions to find the power series expansion of $$\frac{1+5x}{1-2x-3x^2}$$

Answer 1

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.

Answer 2

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 $