# Confirmation needed .

• Feb 23rd 2009, 04:01 AM
thereddevils
Confirmation needed .
Expand $\frac{5x}{(1-2x)(2+x)}$ in ascending powers of x up to and including the term in x^3

I tried :

After partial fracting , i got $\frac{1}{1-2x}-\frac{2}{2+x}$

$=(1-2x)^{-1}-4(1+\frac{x}{2})^{-1}$

$=1+2x+4x^2+8x^3-4+2x-x^2+\frac{1}{2}x^3$

$=-3+4x+3x^2+\frac{17}{2}x^3$

which is wrong according to the book .
• Feb 23rd 2009, 04:06 AM
Power series
Hello thereddevils
Quote:

Originally Posted by thereddevils
Expand $\frac{5x}{(1-2x)(2+x)}$ in ascending powers of x up to and including the term in x^3

I tried :

After partial fracting , i got $\frac{1}{1-2x}-\frac{2}{2+x}$

$=(1-2x)^{-1}-\color{red}4\color{black}(1+\frac{x}{2})^{-1}$

$=1+2x+4x^2+8x^3-4+2x-x^2+\frac{1}{2}x^3$

$=-3+4x+3x^2+\frac{17}{2}x^3$

which is wrong according to the book .

$\frac{2}{2+x}= \color{red}1\color{black}(1+\frac{x}{2})^{-1}$

• Feb 23rd 2009, 04:28 AM
thereddevils
oh thanks ... so is my working before the mistake correct ?
• Feb 23rd 2009, 05:53 AM
Power series
Hello thereddevils
Quote:

Originally Posted by thereddevils
oh thanks ... so is my working before the mistake correct ?

Yes.

$\frac{5x}{2}+\frac{15x^2}{4}+\frac{65x^3}{8}$