Power Series of $\frac{3}{(1-3x)^2}$ The problem is to find the power series of this function $$\frac{3}{(1-3x)^2}$$ centered at $x = 0$. 
Normally you convert it into $\frac{1}{1-x}$ form. Since the denominator is squared do you multiply that out first then convert it into standard form?


Using the standard Taylor approach, compute the successive derivatives:
$$f(x)=\frac3{(1-3x)^2},f(0)=3$$
$$f'(x)=\frac{3\cdot3\cdot2(1-3x)}{(1-3x)^4}=\frac{3\cdot3\cdot2}{(1-3x)^3},f'(0)=3\cdot3\cdot2$$
$$f''(x)=\frac{3\cdot3^2\cdot3\cdot2(1-3x)^2}{(1-3x)^6}=\frac{3\cdot3^2\cdot3\cdot2}{(1-3x)^4},f''(0)=3\cdot3^2\cdot3\cdot2$$
$$f'''(x)=\frac{3\cdot3^3\cdot4\cdot3\cdot2(1-3x)^3}{(1-3x)^8}=\frac{3\cdot3^3\cdot4\cdot3\cdot2(1-3x)^3}{(1-3x)^5},f'''(0)=3\cdot3^3\cdot4\cdot3\cdot2$$
$$\cdots$$
$$f^{(n)}(0)=3\cdot3^n\cdot(n+1)!$$
So the series is$$\sum_{n=0}^\infty3\cdot3^n\cdot(n+1)\cdot x^n.$$
There is no essential difference with the form $1/(1-x)$.
