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

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Question

(b) In ${\left( {{x^2} - \frac{1}{{3x}}} \right)^9},$

${T_{r + 1}} = {\,^9}{C_r}{({x^2})^{9 - r}}{\left( { - \frac{1}{{3x}}} \right)^r}$

$ = {\,

Vedclass · Accepted Answer

$\frac{{28}}{{243}}$

Vedclass · Answer

(b) In ${\left( {{x^2} - \frac{1}{{3x}}} ight)^9},$

${T_{r + 1}} = {\,^9}{C_r}{({x^2})^{9 - r}}{\left( { - \frac{1}{{3x}}} ight)^r}$

$ = {\,^9}{C_r}{x^{18 - 2r}}\frac{{{{( - 1)}^r}}}{{{3^r}}}{x^{ - r}}$ 

It is independent of $x$. 

$	herefore  18 - 3r = 0 \Rightarrow r = 6$ 

 $	herefore {T_7} = {\,^9}{C_6}{x^{18 - 12}}\frac{{{{( - 1)}^6}}}{{{3^6}}}{x^{ - 6}} = {\,^9}{C_6}\frac{{{{( - 1)}^6}}}{{36}} = \frac{{28}}{{243}}$

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

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