Question

If the coefficients of $x^{-2}$ and $x^{-4}$ in the expansion of ${\left(x^{\frac{1}{3}}+\frac{1}{2x^{\frac{1}{3}}}\right)}^{18},(x>0)$, are $m$ and $n$ respectively , then $\frac{m}{n}$ is equal to :

• A. 182
• B. $\frac{4}{5}$
• C. $\frac{5}{4}$
• D. 27

Answer 1

Answer: (A)

$T_{r+1}={}^{18}{C}_r{\left(x^{1/3}\right)}^{18-r}{\left(\frac{1}{2x^{1/3}}\right)}^r$
$={}^{18}{C}_r{\left(\frac{1}{2}\right)}^r{x}^{\frac{18-2r}{3}}$
For coefficient of $x^{-2},\frac{18-2r}{3}=-2$
$\Rightarrow r=12$
For coefficient of $x^{-4},\frac{18-2r}{3}=-4$
$\Rightarrow r=15$
$\Rightarrow \frac{m}{n}=\frac{{}^{18}{C}_{12}{\left(\frac{1}{2}\right)}^{12}}{{}^{18}{C}_{15}{\left(\frac{1}{2}\right)}^{15}}$
$\frac{{}^{18}{C}_8(2{)}^3}{{}^{18}{C}_3}=182$