JEE Main 2016 · 10 Apr · Shift 1 · Mathematics Q67

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MathematicsSingle correct+4 / -1Question 67 of 90

If the coefficients of $x^{-2}$ and $x^{-4}$ in the expansion of $\left(x^{1/3}+\dfrac1{2x^{1/3}}\right)^{18}$ , (x > 0), are m and n respectively, then $\dfrac mn$ is equal to :

182

\dfrac45

\dfrac54

Answer and worked solution

Official answer: A

$T_{r+1}=\binom{18}r2^{-r}x^{(18-2r)/3}$ . $x^{-2}$ needs r = 12 and $x^{-4}$ needs r = 15. So
$\frac mn=\frac{\binom{18}{12}2^{-12}}{\binom{18}{15}2^{-15}}=8\cdot\frac{18564}{816}=182.$

Topics:General and middle terms

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JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 67