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)

$T _{ r +1}={ }^{18} C _{ r }\left( x ^{1 / 3}\right)^{18- r }\left(\frac{1}{2 x ^{1 / 3}}\right)^{ r }$
$={ }^{18} C _{ r }\left(\frac{1}{2}\right)^{ r } x ^{\frac{18-2 r }{3}}$
For coefficient of $x^{-2}, \frac{18-2 r}{3}=-2$
$\Rightarrow r=12$
For coefficient of $x^{-4}, \frac{18-2 r}{3}=-4$
$\Rightarrow r=15$
$\Rightarrow \frac{m}{n}=\frac{{ }^{18} C_{12}\left(\frac{1}{2}\right)^{12}}{{ }^{18} C_{15}\left(\frac{1}{2}\right)^{15}}$
$\frac{{ }^{18} C _{8}(2)^{3}}{{ }^{18} C _{3}}=182$