MathematicsQ51–51 of 91 questions
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51
MathematicsDifficultMCQJEE Main · 2016
If the coefficients of $x^{-2}$ and $x^{-4}$ in the expansion of ${\left( {{x^{\frac{1}{3}}} + \frac{1}{{2{x^{\frac{1}{3}}}}}} \right)^{18}}, (x > 0),$ are $m$ and $n$ respectively,then $\frac{m}{n}$ is equal to
A
$27$
B
$182$
C
$\frac{5}{4}$
D
$\frac{4}{5}$
Solution
(B) The general term in the expansion is $T_{r+1} = ^{18}C_{r} (x^{1/3})^{18-r} (\frac{1}{2x^{1/3}})^r$.
Simplifying this,we get $T_{r+1} = ^{18}C_{r} \cdot \frac{1}{2^r} \cdot x^{6 - r/3 - r/3} = ^{18}C_{r} \cdot \frac{1}{2^r} \cdot x^{6 - 2r/3}$.
For the coefficient of $x^{-2}$,set $6 - \frac{2r}{3} = -2$,which gives $\frac{2r}{3} = 8$,so $r = 12$.
Thus,$m = ^{18}C_{12} \cdot \frac{1}{2^{12}}$.
For the coefficient of $x^{-4}$,set $6 - \frac{2r}{3} = -4$,which gives $\frac{2r}{3} = 10$,so $r = 15$.
Thus,$n = ^{18}C_{15} \cdot \frac{1}{2^{15}}$.
Now,$\frac{m}{n} = \frac{^{18}C_{12} \cdot 2^{-12}}{^{18}C_{15} \cdot 2^{-15}} = \frac{^{18}C_{12}}{^{18}C_{15}} \cdot 2^3$.
Using $^{18}C_{12} = ^{18}C_{6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 18564$ and $^{18}C_{15} = ^{18}C_{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816$.
$\frac{m}{n} = \frac{18564}{816} \times 8 = \frac{18564}{102} = 182$.
Simplifying this,we get $T_{r+1} = ^{18}C_{r} \cdot \frac{1}{2^r} \cdot x^{6 - r/3 - r/3} = ^{18}C_{r} \cdot \frac{1}{2^r} \cdot x^{6 - 2r/3}$.
For the coefficient of $x^{-2}$,set $6 - \frac{2r}{3} = -2$,which gives $\frac{2r}{3} = 8$,so $r = 12$.
Thus,$m = ^{18}C_{12} \cdot \frac{1}{2^{12}}$.
For the coefficient of $x^{-4}$,set $6 - \frac{2r}{3} = -4$,which gives $\frac{2r}{3} = 10$,so $r = 15$.
Thus,$n = ^{18}C_{15} \cdot \frac{1}{2^{15}}$.
Now,$\frac{m}{n} = \frac{^{18}C_{12} \cdot 2^{-12}}{^{18}C_{15} \cdot 2^{-15}} = \frac{^{18}C_{12}}{^{18}C_{15}} \cdot 2^3$.
Using $^{18}C_{12} = ^{18}C_{6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 18564$ and $^{18}C_{15} = ^{18}C_{3} = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} = 816$.
$\frac{m}{n} = \frac{18564}{816} \times 8 = \frac{18564}{102} = 182$.
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