Question

When $|x | < \frac{1}{2} ,$ the coefficient of $x^4$ in the expansion of $\frac{3x^{2} - 5x+3}{\left(x-1\right)\left(2x+1\right)\left(x+3\right)}$ is

• A. $\frac{722}{27}$
• B. $\frac{724}{27}$
• C. $\frac{ - 722}{27}$
• D. $\frac{-724}{27}$

Answer 1

Answer: (C)

Given expression $\frac{3 x^{2}-5 x+3}{(x-1)(2 x+1)(x+3)}$
$=-\left(x^{2}-\frac{5}{3} x+1\right)(1-x)^{-1}(1+2 x)^{-1}\left(1+\frac{x}{3}\right)^{-1}$
$=-\left(x^{2}-\frac{5}{3} x+1\right)\left(1+x+x^{2}+x^{3}+x^{4}\right) $
$\left(1-2 x+4 x^{2}-8 x^{3}+16 x^{4}\right) $
$\left(1-\frac{x}{3}+\frac{x^{2}}{9}-\frac{x^{3}}{27}+\frac{x^{4}}{81}\right) $
[On neglecting higher degree terms]
$=-\left[1+x+x^{2}+x^{3}+x^{4}-\left(\frac{5}{3} x+\frac{5}{3} x^{2}\right.\right.$
$\left.\left.+\frac{5}{3} x^{3}+\frac{5}{3} x^{4}\right)+x^{2}+x^{3}+x^{4}\right]$
$\left[1-\frac{x}{3}+\frac{x^{2}}{9}-\frac{x^{3}}{27}+\frac{x^{4}}{81}-2 x+\frac{2}{3} x^{2}\right.$
$-\frac{2}{9} x^{3}+\frac{2}{27} x^{4}+4 x^{2}-\frac{4}{3} x^{3}+\frac{4}{9} x^{4}$
$\left.-8 x^{3}+\frac{8}{3} x^{4}+16 x^{4}\right]$
$=-\left[1-\frac{2}{3} x+\frac{x^{2}}{3}+\frac{x^{3}}{3}+\frac{x^{4}}{3}\right]$
$\left[1-\frac{7 x}{3}+\frac{43}{9} x^{2}-\frac{259}{27} x^{3}+\frac{1555}{81} x^{4}\right]$
So, coefficient of $x^{4}$ in the expansion of
$\frac{3 x^{2}-5 x+3}{(x+1)(2 x+1)(x+3)}$
$=\left[-\frac{1555}{81}+\frac{518}{81}+\frac{43}{27}-\frac{7}{9}+\frac{1}{3}\right]$
$=-\frac{1555+518+129-63+27}{81}$
$=-\frac{2166}{81}=-\frac{722}{27}$