The term independent of x in {left( {2x - fra | vedclass

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(D) The general term $T_{r+1}$ in the expansion of ${\left( {2x - \frac{1}{{2{x^2}}}} \right)^{12}}$ is given by:$T_{r+1} = {^{12}C_r} {(2x)^{12-r}} {

Vedclass · Accepted Answer

$7920$

Vedclass · Answer

(D) The general term $T_{r+1}$ in the expansion of ${\left( {2x - \frac{1}{{2{x^2}}}} ight)^{12}}$ is given by:$T_{r+1} = {^{12}C_r} {(2x)^{12-r}} {\left( -\frac{1}{2x^2} ight)^r}$$T_{r+1} = {^{12}C_r} {2^{12-r}} {x^{12-r}} {(-1)^r} {2^{-r}} {x^{-2r}}$$T_{r+1} = {^{12}C_r} {2^{12-2r}} {(-1)^r} {x^{12-3r}}$For the term to be independent of $x$,the power of $x$ must be $0$:$12 - 3r = 0 \Rightarrow r = 4$Substituting $r = 4$ into the expression:$T_{4+1} = {^{12}C_4} {2^{12-8}} {(-1)^4} = {^{12}C_4} {2^4}$$T_5 = \frac{12 	imes 11 	imes 10 	imes 9}{4 	imes 3 	imes 2 	imes 1} 	imes 16 = 495 	imes 16 = 7920$.

The term independent of $x$ in ${\left( {2x - \frac{1}{{2{x^2}}}} \right)^{12}}$ is

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