The term independent of x in the expansion of | vedclass

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(B) The given expression is $\left( \frac{1}{60} - \frac{x^8}{81} \right) \left( 2x^2 - \frac{3}{x^2} \right)^6$.The general term $T_{r+1}$ in the exp

Vedclass · Accepted Answer

$-36$

Vedclass · Answer

(B) The given expression is $\left( \frac{1}{60} - \frac{x^8}{81} ight) \left( 2x^2 - \frac{3}{x^2} ight)^6$.The general term $T_{r+1}$ in the expansion of $\left( 2x^2 - \frac{3}{x^2} ight)^6$ is given by:$T_{r+1} = {^6C_r} (2x^2)^{6-r} \left( -\frac{3}{x^2} ight)^r = {^6C_r} 2^{6-r} (-3)^r x^{12-2r-2r} = {^6C_r} 2^{6-r} (-3)^r x^{12-4r}$.To find the term independent of $x$ in the product $\left( \frac{1}{60} - \frac{x^8}{81} ight) \left( 2x^2 - \frac{3}{x^2} ight)^6$,we need:$1$. The term independent of $x$ from the expansion of $\left( 2x^2 - \frac{3}{x^2} ight)^6$,which occurs when $12-4r = 0 \Rightarrow r = 3$.For $r=3$,the term is ${^6C_3} 2^{6-3} (-3)^3 = 20 	imes 8 	imes (-27) = -4320$.Multiplying by $\frac{1}{60}$,we get $\frac{1}{60} 	imes (-4320) = -72$.$2$. The term containing $x^{-8}$ from the expansion of $\left( 2x^2 - \frac{3}{x^2} ight)^6$,which occurs when $12-4r = -8$ $\Rightarrow 4r = 20$ $\Rightarrow r = 5$.For $r=5$,the term is ${^6C_5} 2^{6-5} (-3)^5 = 6 	imes 2 	imes (-243) = -2916$.Multiplying by $-\frac{x^8}{81}$,we get $-\frac{1}{81} 	imes (-2916) = 36$.Adding these results,the total term independent of $x$ is $-72 + 36 = -36$.Thus,the correct answer is option $(B)$.

The term independent of $x$ in the expansion of $\left( \frac{1}{60} - \frac{x^8}{81} \right) \left( 2x^2 - \frac{3}{x^2} \right)^6$ is equal to

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