# Manipulating formal power series help please

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

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

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

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

-

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
show 7 more comments

## 1 Answer

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).

-

Oh ~ that 2nd line makes alot more sense now. – Jono May 4 '11 at 20:50
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9269309639930725, "perplexity_flag": "middle"}