The term independent of x in the expansion {l | vedclass

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

Question

(B) The general term in the expansion of ${\left( {{x^2} - \frac{1}{{3x}}} \right)^9}$ is given by ${T_{r + 1}} = {\,^9}{C_r}{({x^2})^{9 - r}}{\left(

Vedclass · Accepted Answer

$\frac{28}{243}$

Vedclass · Answer

(B) The general term in the expansion of ${\left( {{x^2} - \frac{1}{{3x}}} ight)^9}$ is given by ${T_{r + 1}} = {\,^9}{C_r}{({x^2})^{9 - r}}{\left( { - \frac{1}{{3x}}} ight)^r}$.Simplifying the expression,we get ${T_{r + 1}} = {\,^9}{C_r} \cdot {x^{18 - 2r}} \cdot \frac{{{{( - 1)}^r}}}{{{3^r}}} \cdot {x^{ - r}} = {\,^9}{C_r} \cdot \frac{{{{( - 1)}^r}}}{{{3^r}}} \cdot {x^{18 - 3r}}$.For the term to be independent of $x$,the exponent of $x$ must be $0$,so $18 - 3r = 0$,which gives $r = 6$.Substituting $r = 6$ into the general term,we get ${T_7} = {\,^9}{C_6} \cdot \frac{{{{( - 1)}^6}}}{{{3^6}}} = \frac{{9 	imes 8 	imes 7}}{{3 	imes 2 	imes 1}} \cdot \frac{1}{{729}} = 84 \cdot \frac{1}{{729}} = \frac{{28}}{{243}}$.

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

Similar Questions

If the coefficients of $(r-5)^{th}$ and $(2r-1)^{th}$ terms in the expansion of $(1+x)^{34}$ are equal,find $r$.

If the term independent of $x$ in the expansion of $(\sqrt{a}x^2 + \frac{1}{2x^3})^{10}$ is $105$,then $a^2$ is equal to:

The coefficient of $x^5$ in the expansion of $(1 + x^2)^5(1 + x)^4$ is

The terms containing $x^r y^s$ (for certain $r$ and $s$) are present in both the expansions of $(x+y^2)^{13}$ and $(x^2+y)^{14}$. If $\alpha$ is the number of such terms,then the sum $\alpha \sum_{r, s}(r+s) =$

The coefficient of $x^{-5}$ in the binomial expansion of ${\left( {\frac{{x + 1}}{{{x^{2/3}} - {x^{1/3}} + 1}} - \frac{{x - 1}}{{x - {x^{1/2}}}}} \right)^{10}}$ where $x \ne 0, 1$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module