If the term independent of x in the expansion | vedclass

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(D) The general term in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^r$.For the expansion $\left(\frac{3}{2} x^{2}-\frac{1}

Vedclass · Accepted Answer

$7$

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(D) The general term in the expansion of $(a+b)^n$ is given by $T_{r+1} = {}^{n}C_{r} a^{n-r} b^r$.For the expansion $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}ight)^{9}$,the general term is:$T_{r+1} = {}^{9}C_{r} \left(\frac{3}{2} x^{2}ight)^{9-r} \left(-\frac{1}{3x}ight)^{r}$$T_{r+1} = {}^{9}C_{r} \left(\frac{3}{2}ight)^{9-r} \left(-\frac{1}{3}ight)^{r} x^{18-2r} x^{-r}$$T_{r+1} = {}^{9}C_{r} \left(\frac{3}{2}ight)^{9-r} \left(-\frac{1}{3}ight)^{r} x^{18-3r}$For the term to be independent of $x$,the exponent of $x$ must be $0$:$18 - 3r = 0 \implies r = 6$Substituting $r = 6$ to find $k$:$k = {}^{9}C_{6} \left(\frac{3}{2}ight)^{9-6} \left(-\frac{1}{3}ight)^{6}$$k = {}^{9}C_{3} \left(\frac{3}{2}ight)^{3} \left(\frac{1}{3}ight)^{6}$$k = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} 	imes \frac{27}{8} 	imes \frac{1}{729}$$k = 84 	imes \frac{27}{8} 	imes \frac{1}{729} = 84 	imes \frac{1}{8 	imes 27} = \frac{84}{216} = \frac{7}{18}$Therefore,$18k = 18 	imes \frac{7}{18} = 7$.

If the term independent of $x$ in the expansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ is $k,$ then $18 k$ is equal to

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