Question

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

Answer 1

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

Then,

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

On equating coefficients, we get

$a_0=\frac{3}{2}$, $2a_1-a_0=1$

Whence $a_1=\frac{5}{4}$

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

$\therefore$ $2a_2=\frac{11}{4}$

$a_2=\frac{11}{8}$

Also ,

$2a_3=\frac{11}{8}+\frac{5}{4}$ i.e. $a_3=\frac{21}{16}$

Hence ,

$\frac{3+x}{2-x-x^2}=\frac{3}{2}+\frac{5}{4}x+\frac{11}{8}{x}^2+\frac{21}{16}{x}^3+....$