Computing the coefficient of $x^n$ in the following expansion The coefficient of $x^{-n}$ in the expansion of $\frac{2-3x}{1-3x+2x^2}$ is 
$a.)$ $(-3)^n - (2)^{\frac{1}{2}n -1} $
$b.)$ $2^n + 1 $
$c.)$ $ 3(2)^{\frac{1}{2}n - 1} - 2(3)^n $
$d.)$ None of the foregoing numbers. 
My attempt :
First of all I concluded that the question must be asking about coeff. to the power of -n , not n as mentioned because x only appears in the denominator.
I was able to reduce the given expression to the following separated fractions : $ \frac{1}{1-2x} $ and $ \frac{1}{1-x} $. Since both the terms contain x it must be the case that the total power of each term in the binomial expansion must be -n. 
Hence my answer turned out to be $2^n$. The correct option as indicated is option b.)
Please tell me where am I going wrong in my method, and suggest a better method to solve the following problem.  
 A: Partial fraction decomposition is a proper method to answer the question. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.

We obtain
  \begin{align*}
[x^n]\frac{2-3x}{1-3x+2x^2}&=[x^n]\left(\frac{1}{1-2x}+\frac{1}{1-x}\right)\tag{1}\\
&=[x^n]\sum_{j=0}^{\infty}(2x)^j+[x^n]\sum_{j=0}^{\infty}x^j\tag{2}\\
&=2^n+1
\end{align*}

Comment:


*

*In (1) we use partial fraction decomposition

*In (2) we use the linearity of the coefficient of operator and the geometric series expansion from which the coefficient of $x^n$ can easily be obtained.
Note: Since no explicit statement about the center of series expansion is formulated, we may assume an expansion around $x=0$ is expected.
A: You are correct in your division of the polynomials. They are asking for the taylor coefficients in the power series expansion of the function. Let $f(x) = \frac{1}{1-2x}$ and $g(x) = \frac{1}{1-x}$. Then
\begin{align*}
    f'(x) = \frac{2}{(1-2x)^2}\\
    f''(x) = \frac{2\times 2^2}{(1-2x)^3}\\
    f^{(3)}(x) = \frac{6\times 2^3}{(1-2x)^4}\\
    f^{(n)}(x) = \frac{n!\times 2^n}{(1-2x)^{n+1}}\\
\end{align*}
and similarly you can see that $g^{(n)}(x) = \frac{n!}{(1-x)^{n+1}}$. Expanding the function around $0$ will give you the right answer; your function is $F(x) = f(x) + g(x)$ so that 
    $$F(x) = \sum_{n=0}^{\infty}\frac{F^{(n)}(0)}{n!}x^n = \sum_{n=0}^{\infty}\frac{f^{(n)}(0) + g^{(n)}(0)}{n!}x^n$$
