# Confirmation needed .

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• Feb 23rd 2009, 04:01 AM
thereddevils
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 .
• Feb 23rd 2009, 04:06 AM
Grandad
Power series
Hello thereddevils
Quote:

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 .

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

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

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

Yes.

I make the final answer

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

Does that agree with your book?

Grandad
• Feb 23rd 2009, 07:02 AM
thereddevils
Absolutely yes , thanks for helping out .