Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

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

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

share|improve this question
    
It is not clear what $m$ is. Multiply both sides and show us your calculation. –  Phira May 4 '11 at 18:51
    
It is rather hard to see what you mean by your third sentence "When i tried...". –  Mariano Suárez-Alvarez May 4 '11 at 18:53
    
Working out shown –  Jono May 4 '11 at 19:01
    
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. –  Phira May 4 '11 at 19:12
    
@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. –  Phira May 4 '11 at 19:14

1 Answer 1

up vote 2 down vote accepted

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).

share|improve this answer
    
Oh ~ that 2nd line makes alot more sense now. –  Jono May 4 '11 at 20:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.