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