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

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(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

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