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 $