The coefficient of $x^n$ in the expansion of $\frac{2-3x}{1-3x+2x^2}$ is? The coefficient of $x^n$ in the expansion of $\frac{2-3x}{1-3x+2x^2}$ is?
Working: $1 – 3x + 2x^2 = (1 – x)(1 – 2x)
=> (1 – x)^{-1} = 1 + (-1)(-x) + {(-1)(-1 -1)/2!}(-x)^2 + {(-1)(-1-1)(-12)/3!}(-x)^3 + … = 1 + x + x^2 + x^3 + …. $
$(1 – 2x)^{-1} = 1 + (-1)(-2x) + {(-1)(-1 – 1)/2!}(-2x)^2 + {(-1)(-1-1)(1-2)/3!}(-2x)^3 + … = 1 + 2x + (2x)^2 + (2x)^3 + …..$
Coefficient of $x^n$ in $(1 – x)^{-1}(1 – 2x)^{-1} = 2^n + 2^{n-1} + 2^{n-2} + …. + 2 + 1 = 2^{n+1} – 1$
Coefficient of $x^{n-1}$ in $(1 – x)^{-1}(1 – 2x)^{-1}$ = $2^n – 1$
Coefficient of $x^n$ in $(2 – 3x)(1 – x)^{-1}(1 – 2x)^{-1} = 2^{n+2} – 2 – 3(2n – 1)$ = $2^{n+2} – 3(2)^n + 1 = (4)2^n – (3)2^n + 1 = 2^n + 1.$
Is there a shorter, quicker (and easier) method to do this?

$\dfrac{2-3x}{1-3x+2x^2}=\dfrac{1}{1-x}+\dfrac1{1-2x}$
The coefficient of $x^n$ is $1^n+2^n=2^n+1$
