1
$\begingroup$

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?

$$\frac1{(1+3x)^3} = 1 - 9x + \cdots + \dfrac{n(n-1)}{18}(-3)^nx^{n-2}+\cdots $$

  • 2
    To divide a power series by $18$ you divide each term by $18$. So yes, exactly what you just did.2011-05-04
  • 0
    Thanks Qiaochu, then if i put it into sigma notation, is it infinite to n=0 or infinite to n=2 , since n(0) and n(1) gives 02011-05-04
  • 1
    It doesn't matter.2011-05-04

1 Answers 1