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Say i got $\displaystyle{\frac{(1-2x)}{(1+3x)^3}}$

I used $\displaystyle{\frac{1}{(1+3x)}}$ $=\sum_{n=0}^\infty(-3)^n x^n$ and differentiated twice

I got $\displaystyle{\frac{(1-2x)}{(1+3x)^3}}$ = $=\sum_{n=0}^\infty [((n)(n-1)(-3)^n)/18] x^{(n-2)}$

Multiply (1-2x) on both side I got

= $\sum_{n=0}^\infty [((n)(n-1)(-3)^n)/18] x^{(n-2)}$ - $2\sum_{n=0}^\infty [((n-1)(n-2)(-3)^{(n-1)}))/18] x^{(n-2)}$

$=\sum_{n=0}^\infty [(5n-4)(n-1)(-3)^n /54 ] x^{(n-2)}$

Is that correct ? I had a feeling that its wrong...

Coefficient of $z^{(n-2)}$ is $\displaystyle{\frac{(5n-4)(n-1)(-3)^n}{54}}$?

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    Please write (1-2x)/(1+3x)^3 instead of 1-2x/(1+3x)^3, because a-b/c is likely to be interpreted as a-(b/c) rather than (a-b)/c. Even better, you can use LaTeX here to write $\displaystyle{\frac{1-2x}{(1+3x)^3}}$. The syntax I used for that is `$\displaystyle{\frac{1-2x}{(1+3x)^3}}$`; you put the LaTeX math between dollar signs. Please include the sums when you mean a series. For example, you really want $\frac{1}{1+3x}=\sum_{n=0}^\infty(-3)^nx^n$ rather than what you wrote in the second line. (If you're new to LaTeX, you can right-click then click "Show Source" to see how it is done.)2011-05-04
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    Thank you and I will edit it.2011-05-04
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    @Jono: Thanks. Some of the things in your edit are presumably not what was intended. For example, you now have $1-2x$ in two numerators where I believe you want $1$. The syntax `$\frac{a}{b}$` gives $\frac{a}{b}$, so you can adjust the numerator and denominator in brackets to get the fraction you want.2011-05-04
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    Format adjusted, thank you.2011-05-04
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    When you want exponents of more than one character, you need to put them in braces: x^{(n+1)} gives $x^{(n+1)}$ while x^(n+1) gives $x^(n+1)$2011-05-04
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    All format is corrected. thank you2011-05-04

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