Question

Find the general term of the following expressions when expanded in ascending powers of $x$.
$\frac{4+7x}{(2+3x){(1+x)}^2}$

Answer 1

$\frac{4+7x}{(2+3x)(1+x{)}^2}$

By partial faction we will break it

$\frac{A}{2+3x}+\frac{8x+c}{(1+x)2}=\frac{4+7x}{(2+8x)(1+x^2)}$

$\Rightarrow (A+3B)x^2+(2A+2B+3C)x+A+2C=4+7x$

$A+3B=0...\left(i\right)$

$2A+2B+3C=7...\left(ii\right)$

$A+2C=4...\left(iii\right)$

From (i) + (ii) + (iii)

A = -6

B = 2

C = 5

So

$\frac{-6}{2+3x}+\frac{2x+5}{(1+x{)}^2}$

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

$\Rightarrow -3(1+\frac{3}{2}x{)}^{-1}+(2x+5\left)\right(1+x)-2$

$=-3(1-3/2x+3/4x^2-\frac{27}{3}{x}^3+...)+(2x+5)(1-2x+3x^2-4x^3+...)$

$=(-3+\frac{9}{2}x-\frac{27}{4}{x}^2-\frac{31}{3}{x}^3+...)+(2x-4x^2+6x^3-3x+\mathrm{1...})$

$+(5-10x+15x^2-20x^3+...)$

$=2-\frac{7}{2}x+\frac{17}{4}{x}^2-\frac{193}{8}{x}^3+...$