Question

Find the general term of $\frac{3x^2+x-2}{{(x-2)}^2(1-2x)}$ when expanded in a series of ascending powers of $x$.

Answer 1

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

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

Take $1-2x=0\Rightarrow A=-\frac{1}{3}$

Take $x-2=0\Rightarrow C=-4$

And then $B=-\frac{5}{3}$

We have
$\frac{3x^2+x-2}{{(x-2)}^2(1-2x)}=-\frac{1}{3(1-2x)}-\frac{5}{3(x-2)}-\frac{4}{{(x-2)}^2}$
$=-\frac{1}{3}{(1-2x)}^{-1}+\frac{5}{6}{(1-\frac{x}{2})}^{-1}-{(1-\frac{x}{2})}^{-2}$
Hence the general term of the expansion is
$(-\frac{2^r}{3}+\frac{5}{6}.\frac{1}{2^r}-\frac{r+1}{2^r})x^r$