$\displaystyle f(x)=\frac{3}{2x-1}$ , c=2
This is what I did:
$\displaystyle \frac{a}{1-r} = \frac{-3}{1-2x} = -3\sum^\infty_ {n=0} (2x)^n$
Now, what do I do next since its centered at c=2??
question: what form for the power series are you trying to use? you are trying to use the fact that $\displaystyle \frac 1{1 - x} = \sum_{n = 0}^\infty x^n$, for $\displaystyle |x| < 1$, am i right?
now look at what we have. does it look like it is in that form? is the constant in the denominator a 1? is it 1 - (some function of x) in the denominator? no it isn't. get it into that form before you make the power series
you could start by factoring out -3 from the denominator and then you can just remove it and write -1/3 in front of the fraction. you should review factoring
$\displaystyle \frac {-3}{-3 -2(x - 2)} = \frac {-3}{-3 \left[ 1 + \frac {2(x - 2)}3\right]} = \frac 1{1 + \frac {2(x - 2)}3} = \frac 1{1 - \left[ - \frac {2(x - 2)}3\right]}$
now we have the required form.
if c = 0 it means you want your function part to look like (x - 0) which happens to just be x. thus, having that x there (possibly times some constant) is good enough. then, you have the constant in front being a "+1", so we have the form. you would NOT change it to 4 - 3 - 2x, because that is getting away from the form you want, which is $\displaystyle \frac 1{1 - f(x)}$, where $\displaystyle f(x)$ is some function in $\displaystyle x$. in general, you would want in this case, $\displaystyle \frac 1{1 - f(x - c)}$