Question

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

Answer 1

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

as $x^3$

$\frac{1+2x}{1-x-x^2}$

Let $\frac{1+2x}{1-x-x^2}=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+......$

$\Rightarrow (1+2x)=\left(1-x-x^2\right)[a_0+a_{1}x+a_2{x}^2+a_3{x}^3+......]$

On comparing coefficients, we have

$a_0=1,a_1-a_0=2$, whence $a_1=3$

The coefficients of higher powers of $x$ are found in succession from

the relation $a_n-a_{n-1}-a_{n-2}=0$;

Hence $a_2=4$ and $a_3=7$

Hence, $\frac{1+2x}{1-x-x^2}=[1+3x+4x^2+7x^3+......]$