Formal Power Series Say I differentiate this twice:
$$\dfrac{1}{1+3x} = 1 - 3x + 9x^2 -\cdots+ (-3)^n x^n+\cdots $$
I got 
$$\dfrac{18}{(1+3x)^3}  = 18 - 162x + \cdots + n\cdot(n-1)(-3)^nx^{n-2}+\cdots$$
If I wanted to get $$\dfrac{1}{(1+3x)^3}  = \cdots$$do I just move the 18 over ? Would that work?
$$\frac1{(1+3x)^3} = 1 - 9x + \cdots + \dfrac{n(n-1)}{18}(-3)^nx^{n-2}+\cdots $$ 

Yes. Note that you can write your series as
$$1/(1+3x)^3 = 1 - 9x + \cdots + \dfrac{n(n-1)}{3\cdot3\cdot2}(-3)^nx^{n-2}+\cdots $$ 
$$1/(1+3x)^3 = 1 - 9x + \cdots + \dfrac{n(n-1)}{ 2}(-3x)^{n-2}+\cdots $$ 
$$1/(1+3x)^3 = 1 - 9x + \cdots +{n \choose2}(-3x)^{n-2}+\cdots $$
or changing the index
$$1/(1+3x)^3 = 1 - 9x + \cdots +{n+2 \choose2}(-3x)^{n}+\cdots $$
In general you can write 
$$\dfrac{1}{(1-x)^{k+1}}=\sum_{n=0}^{\infty}{n+k\choose k}x^n$$
Where $$\displaystyle {n+k\choose k}$$ means $$\dfrac{(n+1)(n+2)\cdots(n+k)}{k!}$$
