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(B) The general term in the expansion of $\left(x^{1/3} + \frac{1}{2}x^{-1/3}\right)^{21}$ is given by $T_{r+1} = {}^{21}C_r (x^{1/3})^{21-r} (\frac{1

Vedclass · Accepted Answer

$408$

Vedclass · Answer

(B) The general term in the expansion of $\left(x^{1/3} + \frac{1}{2}x^{-1/3}\right)^{21}$ is given by $T_{r+1} = {}^{21}C_r (x^{1/3})^{21-r} (\frac{1}{2}x^{-1/3})^r = {}^{21}C_r (\frac{1}{2})^r x^{\frac{21-2r}{3}}$.For $p$,the coefficient of $x^{-3}$:$\frac{21-2r}{3} = -3$ $\Rightarrow 21-2r = -9$ $\Rightarrow 2r = 30$ $\Rightarrow r = 15$.Thus,$p = {}^{21}C_{15} (\frac{1}{2})^{15}$.For $q$,the coefficient of $x^{-5}$:$\frac{21-2r}{3} = -5$ $\Rightarrow 21-2r = -15$ $\Rightarrow 2r = 36$ $\Rightarrow r = 18$.Thus,$q = {}^{21}C_{18} (\frac{1}{2})^{18}$.Now,$\frac{5p}{4q} = \frac{5 \cdot {}^{21}C_{15} (\frac{1}{2})^{15}}{4 \cdot {}^{21}C_{18} (\frac{1}{2})^{18}} = \frac{5 \cdot {}^{21}C_6}{4 \cdot {}^{21}C_3} \cdot 2^3 = \frac{5 \cdot {}^{21}C_6}{4 \cdot {}^{21}C_3} \cdot 8 = 10 \cdot \frac{21 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 / 6!}{21 \cdot 20 \cdot 19 / 3!} = 10 \cdot \frac{18 \cdot 17 \cdot 16}{6 \cdot 5 \cdot 4} = 10 \cdot (3 \cdot 17 \cdot 8 / 1) = 408$.

If $p$ and $q$ are respectively the coefficients of $x^{-3}$ and $x^{-5}$ in the expansion of $\left(x^{1/3} + \frac{1}{2x^{1/3}}\right)^{21}, x > 0$,then $\frac{5p}{4q} = $

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