Question

If the coefficients of $x^{-2}$ and $x^{-4}$ in the expansion of ${(x^{\frac{1}{3}}+\frac{1}{2x^{\frac{1}{3}}})}^{18},$ $(x>0)$, 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

The correct option is A $182$

Using binomial theorem,

${(x^{\frac{1}{3}}+\frac{1}{2x^{\frac{1}{3}}})}^{18}=\sum _{n=0}^{18}{}^{18}{C}_n{\left(x^{\frac{1}{3}}\right)}^{18-n}{\left(\frac{1}{2x^{\frac{1}{3}}}\right)}^n$

$=\sum _{n=0}^{18}{}^{18}{C}_n{x}^{\frac{18-n}{3}}\cdot \frac{1}{2^n{x}^{n/3}}$

$=\sum _{n=0}^{18}{}^{18}{C}_n{2}^{-n}{x}^{\frac{18-n}{3}-\frac{n}{3}}$

$=\sum _{n=0}^{18}{}^{18}{C}_n{2}^{-n}{x}^{\frac{18-2n}{3}}$

and $T_{r+1}={}^{18}{C}_r{2}^{-r}{x}^{\frac{18-2r}{3}}$

$m=$ coefficient of $x^{-2}={}^{18}{C}_{12}{2}^{-12}=\frac{18!}{12!6!}\cdot 2^{-12}$ $[\frac{18-2r}{3}=-2,18-2r=-6,\Rightarrow 18+6=2r,\Rightarrow r=\frac{24}{2}=12]$

$n=$ coefficient of $x^{-4}={}^{18}{C}_{15}{2}^{-15}=\frac{18!}{15!3!}{2}^{-15}$ $[18-\frac{2r}{3}=-4,\Rightarrow 18-12=-12,\Rightarrow 18+12=2r,\Rightarrow r=\frac{30}{2}]$

$\therefore \frac{m}{n}=\frac{\frac{18!}{12!6!}{2}^{-12}}{\frac{18!}{15!3!}{2}^{-15}}$

$=\frac{15!3!}{12!6!}{2}^{12+15}$

$=\frac{15\times 14\times 13\times 12!\times 3!}{12!\times 6\times 5\times 4\times 3!}{2}^3=\frac{15\times 14\times 13}{6\times 5\times 4}\times 2^3$

$=\frac{91}{4}\times 2^3$

$=182$.