g(x) = 2 / (1-x)^3 center c=0
You haven't done anything that jhevon suggested!
He suggested that you write out the power series for $\displaystyle \frac{1}{1- x}$ and differentiate that twice!
What is the power series for $\displaystyle \frac{1}{1- x}$?
(Hint: the sum of the geometric series $\displaystyle \sum_{n=0}^\infty ar^n$ is $\displaystyle \frac{a}{1- r}$.)
the left side of your third equation is incorrect. and when i said differentiate the power series for 1/(1 - x) i was talking about differentiating the expression behind the summation sign in $\displaystyle \sum_{n = 0}^\infty x^n$, for $\displaystyle |x|<1$, twice
or you can follow awkwards suggestion if you learned the negative binomial theorem