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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 $

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    It doesn't matter.2011-05-04

1 Answers 1

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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 $\dfrac{(n+1)(n+2)\cdots(n+k)}{k!}$