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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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To divide a power series by $18$ you divide each term by $18$. So yes, exactly what you just did. –  Qiaochu Yuan May 4 '11 at 9:42
    
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 0 –  Jono May 4 '11 at 9:47
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It doesn't matter. –  Qiaochu Yuan May 4 '11 at 10:55
    
Thank you very much –  Jono May 4 '11 at 11:02
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1 Answer

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

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