1. ## Confirmation needed .

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

I tried :

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

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

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

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

which is wrong according to the book .

2. ## Power series

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

I tried :

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

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

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

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

which is wrong according to the book .
$\displaystyle \frac{2}{2+x}= \color{red}1\color{black}(1+\frac{x}{2})^{-1}$

3. oh thanks ... so is my working before the mistake correct ?

4. ## Power series

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

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