The term independent of x in the expansion of | vedclass

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

Question

(D) The general term $T_{r+1}$ in the expansion of ${\left( {{x^2} - \frac{1}{x}} \right)^9}$ is given by:$T_{r+1} = {^9C_r} {(x^2)}^{9-r} {(-\frac{1}

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The general term $T_{r+1}$ in the expansion of ${\left( {{x^2} - \frac{1}{x}} ight)^9}$ is given by:$T_{r+1} = {^9C_r} {(x^2)}^{9-r} {(-\frac{1}{x})}^r$$T_{r+1} = {^9C_r} {x^{18-2r}} {(-1)^r} {x^{-r}}$$T_{r+1} = {^9C_r} {(-1)^r} {x^{18-3r}}$For the term to be independent of $x$,the exponent of $x$ must be $0$:$18 - 3r = 0$$3r = 18$$r = 6$Substituting $r = 6$ into the general term:$T_{6+1} = {^9C_6} {(-1)^6} {x^0}$$T_7 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} 	imes 1 = 84$Since $84$ is not among the options $A, B, C$,the correct choice is $D$.

The term independent of $x$ in the expansion of ${\left( {{x^2} - \frac{1}{x}} \right)^9}$ is

Similar Questions

The term independent of $x$ in the expansion of $(1-3x+2x^3)(\frac{3x^2}{2}-\frac{1}{3x})^9$ is

Find the coefficient of $x^{64}$ in the expansion of $P(x) = (x - 1)^2(x - 2)^3(x - 3)^4 \dots (x - 10)^{11}$.

If $\lim _{x \rightarrow 1} \frac{\sqrt[4]{x^{-3}}+a \sqrt[4]{x^5}}{x-1}=-2$,then the coefficient of $x$ in the expansion of $(\sqrt[4]{x^{-3}}+a \sqrt[4]{x^5})^4$ is

The coefficient of $x^{24}$ in the expansion of $(1+x^2)^{12}(1+x^{12})(1+x^{24})$ is

If the ratio of the $7^{th}$ term from the beginning to the $7^{th}$ term from the end in the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$ is $\frac{1}{6}$,then $n=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module