If the term independent of x in the expansion | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$7$

Vedclass · Answer

(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

Similar Questions

If the coefficient of the $4^{th}$ term in the expansion of $(a + b)^n$ is $56$,then $n$ is

If the coefficients of $(2 \alpha+4)$-th and $(\alpha-2)$-th terms in the expansion of $(1+x)^{2018}$ are equal,then $\alpha=$

In the expansion of $(x-1)(x-2) \ldots (x-18)$,the coefficient of $x^{17}$ is:

If $L$ and $M$ are respectively the coefficient of $x^{-7}$ in $\left(a x+\frac{b}{x^2}\right)^{11}$ and the coefficient of $x^7$ in $\left(b x^2+\frac{a}{x}\right)^{11}$,then $L+M=$

The term independent of $x$ in the expansion of $\left( 9x - \frac{1}{3\sqrt{x}} \right)^{18}, x > 0$,is $\alpha$ times the corresponding binomial coefficient. Then $\alpha$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module