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

Let $(x + 10)^{50} + (x - 10)^{50} = a_0 + a_1x + a_2x^2 + .... + a_{50}x^{50}$,for $x \in R$; then $\frac{a_2}{a_0}$ is equal to

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

Let $a_0, a_1, \ldots, a_{23}$ be real numbers such that $(1+\frac{2}{5} x)^{23} = \sum_{i=0}^{23} a_i x^i$ for every real number $x$. Let $a_r$ be the largest among the numbers $a_j$ for $0 \leq j \leq 23$. Then the value of $r$ is $....$ .

If the second,third,and fourth terms in the expansion of $(x + a)^n$ are $240, 720,$ and $1080$ respectively,then the value of $n$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module