The term independent of x in the expansion of | vedclass

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

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None of these

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(D) The general term $T_{r+1}$ in the expansion of ${\left( {{x^2} - \frac{1}{x}} ight)^9}$ is given by:$T_{r+1} = {^9C_r} {(x^2)}^{9-r} {(-\frac{1}{x})}^r$$T_{r+1} = {^9C_r} {x^{18-2r}} {(-1)^r} {x^{-r}}$$T_{r+1} = {^9C_r} {(-1)^r} {x^{18-3r}}$For the term to be independent of $x$,the exponent of $x$ must be $0$:$18 - 3r = 0$$3r = 18$$r = 6$Substituting $r = 6$ into the general term:$T_{6+1} = {^9C_6} {(-1)^6} {x^0}$$T_7 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} 	imes 1 = 84$Since $84$ is not among the options $A, B, C$,the correct choice is $D$.

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

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