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

If i got up to this, how could i get $\displaystyle{\frac{(1-2x)}{(1+3x)^3}}$ ?

When i tried to multiply both side, some people says n-m some says n+m for $x^m$

Could someone kindly show me the working out please?

My working out:

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

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

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

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

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    It is not clear what $m$ is. Multiply both sides and show us your calculation.2011-05-04
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    It is rather hard to see what you mean by your third sentence "When i tried...".2011-05-04
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    Working out shown2011-05-04
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    You are not taking care of the start of the sum range. I strongly suggest to not use a formula "n+m", but to actually multiply by $x$ and then shift the summation range by substituting $k-1$ for $n$. This will show you how to modify the summation range and the terms.2011-05-04
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    @Shai It is possible to interpret the first equation properly, but I agree that this is one more instance of not paying attention to the summation range.2011-05-04
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    @user9325: I have just noticed that myself.2011-05-04
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    Could you please show me some working out cause I been struggling with this question for awhile now =\2011-05-04
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    Seriously I been stuck on this question for 3 hours... Someone please help me out ?2011-05-04
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    @Jono: OK, hopefully I will help you soon.2011-05-04
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    sorry, and i really have to head to class in 3 hours time, its 6am here =\ i really want to solve this in order to solve my next two questions before i go to class.2011-05-04
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    Jono: was there a reason to repost you question? See http://math.stackexchange.com/questions/36951/generating--explicit-formula. I realize you're feeling stuck and frantic (putting off work, or putting off getting help tends to do that), but it's best to post a question only once. If you've revised something, or made progress, just edit the original question.2011-05-04

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So we are given the (interesting) equality $$ \frac{{18}}{{(1 + 3x)^3 }} = \sum\limits_{n = 2}^\infty {n(n - 1)( - 3)^n x^{n - 2} } , $$ for $x$ in a neighborhood of $0$. Hence, $$ \frac{{18}}{{(1 + 3x)^3 }} = \sum\limits_{n = 0}^\infty {(n + 2)(n + 1)( - 3)^{n + 2} x^n } , $$ and in turn $$ \frac{1}{{(1 + 3x)^3 }} = \sum\limits_{n = 0}^\infty {\frac{{(n + 2)(n + 1)( - 3)^n }}{2}x^n } . $$ Thus, $$ \frac{{ - 2x}}{{(1 + 3x)^3 }} = \sum\limits_{n = 0}^\infty {\frac{{ - 2(n + 2)(n + 1)( - 3)^n }}{2}x^{n + 1} } = \sum\limits_{n = 1}^\infty {\frac{{ - 2(n + 1)n( - 3)^{n - 1} }}{2}x^n } . $$ Therefore, $$ \frac{{1 - 2x}}{{(1 + 3x)^3 }} = 1 + \sum\limits_{n = 1}^\infty {\frac{{( - 3)^n }}{2}x^n \bigg[(n + 2)(n + 1) + \frac{2}{3}(n + 1)n \bigg]}, $$ or $$ \frac{{1 - 2x}}{{(1 + 3x)^3 }} = 1 + \sum\limits_{n = 1}^\infty {\frac{{( - 3)^n }}{2}\bigg[\frac{5}{3}n^2 + \frac{{11n}}{3} + 2\bigg]x^n } $$ (confirmed numerically).

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    Oh ~ that 2nd line makes alot more sense now.2011-05-04