In the expansion of {left( x - frac{3}{x^2} r | vedclass

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

Vedclass · Accepted Answer

$-2268$

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(D) The general term in the expansion of ${\left( x - \frac{3}{x^2} ight)^9}$ is given by ${T_{r+1}} = {^9C_r} {(x)^{9-r}} {\left( -\frac{3}{x^2} ight)^r}$.Simplifying the expression,we get ${T_{r+1}} = {^9C_r} {(-3)^r} {x^{9-r}} {x^{-2r}} = {^9C_r} {(-3)^r} {x^{9-3r}}$.For the term to be independent of $x$,the exponent of $x$ must be $0$,so $9 - 3r = 0$,which gives $r = 3$.Substituting $r = 3$ into the general term,we get ${T_4} = {^9C_3} {(-3)^3} = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} 	imes (-27) = 84 	imes (-27) = -2268$.

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

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