Question

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

Answer 1

To find the general term of the following expressions when expanded in ascending power of x .

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

Here , we have

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

The Expression can be written as

$\frac{3-2x^2}{(2-3x+x^2{)}^2}=\frac{A}{(2-x{)}^2}+\frac{B}{(2-x)}+\frac{C}{(1-x{)}^2}+\frac{D}{(1-x)}$

$3-2x^2=A(1-x{)}^2+B(2-x\left)\right(1-x{)}^2+C(2-x{)}^2+D(1-x\left)\right(2-x{)}^2$

On Equating the Coefficients , we get

$A+2B+4C+4D=3$

$2B+4D=4$ and

$B=-D$

After Solving for values of $A,B,C$ and $D$

$A=-5,B=-2,C=1,D=2$

Hence , the Expression can be written as

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

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

$\therefore$ Coefficient of $x^r$

$=[-\frac{5}{4}.\left(\frac{r+1}{2^r}\right)-\frac{1}{2^r}+(r+1)+2]x^r$

$=[3+r-\left(\frac{5r+9}{2^{r+1}}\right)]x^r$