Question

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

Answer 1

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

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

On comparing coefficients, we get

$a_0=\frac{1}{2}$ , $2a_1+a_0=1$

Whence $a_1=\frac{1}{4}$

For values of $n>1,$ $2a_n-a_{n-1}-a_{n-2}=0$

$\therefore$ $2a_2=\frac{-3}{4}$

i.e. $a_2=\frac{-3}{8}$

Also, $2a_3=\frac{-3}{8}-\frac{1}{4}$ i.e. $a_3=\frac{1}{16}$

Hence, $\frac{1+x}{2+x+x^2}=\frac{1}{2}+\frac{5}{4}x-\frac{3}{8}{x}^2+\frac{1}{16}{x}^3+....$