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?

 A: Hint: If $\sum_{n=0}^\infty (3x)^n$ is convergent then $$\frac{d}{dx}\left(\sum_{n=0}^\infty (3x)^n\right)=\frac{d}{dx}\left(\frac{1}{1-3x}\right) = \frac{3}{(1-3x)^2}$$
A: Hint: Take the derivative with respect to $x$ of this series 
$$\frac{1}{1-3x} = \sum_{n=0}^\infty (3x)^{n} \, \,\,, |3x|<1$$
A: You can use the Binomial Theorem by writing it as $$3(1-3x)^{-2}$$ This will be valid provided $$|3x|<1$$
A: 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 $1/(1-x)$.
