If the coefficients of x^{-2} and x^{-4} in t | vedclass

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(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}

Vedclass · Accepted Answer

$182$

Vedclass · Answer

(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 $^{18}C_{12} = ^{18}C_{6} = \frac{18 	imes 17 	imes 16 	imes 15 	imes 14 	imes 13}{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1} = 18564$ and $^{18}C_{15} = ^{18}C_{3} = \frac{18 	imes 17 	imes 16}{3 	imes 2 	imes 1} = 816$.$\frac{m}{n} = \frac{18564}{816} 	imes 8 = \frac{18564}{102} = 182$.

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

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