# Thread: expansion of infinite series

1. ## expansion of infinite series

$\displaystyle f(x)=\frac{x^2+5x}{(1+x)(1-x)^2}$

If the expansion of f(x) in ascending power of x is

$\displaystyle c_0+c_1x+c_2x+c_3x+...+c_rx^r+...$

find $\displaystyle c_0,c_1$ and $\displaystyle c_2$ and show that $\displaystyle c_3=11$

2. Originally Posted by thereddevils
$\displaystyle f(x)=\frac{x^2+5x}{(1+x)(1-x)^2}$

If the expansion of f(x) in ascending power of x is

$\displaystyle c_0+c_1x+c_2x+c_3x+...+c_rx^r+...$

find $\displaystyle c_0,c_1$ and $\displaystyle c_2$ and show that $\displaystyle c_3=11$
First notice that

$\displaystyle \frac{x^2+5x}{(1+x)(1-x)^2} = \frac{3}{(1-x)^2} - \frac{2}{1-x} - \frac{1}{1+x}$

If you consider the following geometric power series

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

so

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

$\displaystyle \frac{d}{dx} \left( \frac{1}{1-x}\right) = \frac{1}{(1-x)^2} = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 \cdots$

Substitute above giving

$\displaystyle 3\left( 1 + 2 x + 3 x^2 + 4 x^3\right) - 2 \left( 1 + x + x^2 + x^3\right) - \left( 1 - x + x^2 - x^3\right) \cdots$
$\displaystyle =(3-2-1) + (6 - 2+1)x + (9 - 2-1)x^2 +(12 - 2 +1) x^3 \cdots$
$\displaystyle =5x + 6x^2 + 11 x^3 \cdots$