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(D) The general term in the expansion of $(2x - \frac{3}{x})^6$ is given by $T_{r+1} = {}^6C_r (2x)^{6-r} (-\frac{3}{x})^r$.Simplifying this,we get $T

Vedclass · Accepted Answer

$-4320$

Vedclass · Answer

(D) The general term in the expansion of $(2x - \frac{3}{x})^6$ is given by $T_{r+1} = {}^6C_r (2x)^{6-r} (-\frac{3}{x})^r$.Simplifying this,we get $T_{r+1} = {}^6C_r (2)^{6-r} (-3)^r (x)^{6-r} (x)^{-r} = {}^6C_r (2)^{6-r} (-3)^r (x)^{6-2r}$.For the term to be independent of $x$,the exponent of $x$ must be zero,so $6 - 2r = 0$,which gives $r = 3$.Substituting $r = 3$ into the expression,we get $T_4 = {}^6C_3 (2)^{6-3} (-3)^3$.$T_4 = 20 	imes 2^3 	imes (-27) = 20 	imes 8 	imes (-27) = 160 	imes (-27) = -4320$.

The term independent of $x$ in the expansion of ${\left( {2x - \frac{3}{x}} \right)^6}$ is

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