Question

Expand the following expression in ascending powers of $x$ as far as $x^3$.
$\frac{1}{1+ax-a{x}^2-x^3}$.

Answer 1

To Expand the following expression in ascending powers of x as far

as $x^3$

$\frac{1}{1+ax-a{x}^2-x^3}$

Let the required expansion be

$b_0+b_{1}x+b_2{x}^2+b_3{x}^3+.....$

i.e. $\frac{1}{1+ax-a{x}^2-x^3}$ $=b_0+b_{1}x+b_2{x}^2+b_3{x}^3+.....$

$1=(1+ax-a{x}^2-x^3)[b_0+b_{1}x+b_2{x}^2+b_3{x}^3+.....]$

$\therefore$ On equating coefficients, we get

$b_0=1,b_1+b_{0}a=0$ ; Whence $b_1=-a$

Also, $b_2+b_{1}a-b_{0}a=0$; whence $b_2=a(a+1)$

And $b_3+b_{2}a-b_{1}a-b_0$; whence $b_3=1-2a^2-a^3$

Hence, $\frac{1}{1+ax-a{x}^2-x^3}=1-ax+a(a+1)x^2-(a^3+2a^2-1)x^3+......$