The coefficient of {x^2} in the expansion of | vedclass

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Question

(C) The general term in the expansion of ${\left( x - \frac{1}{2x} \right)^8}$ is given by ${T_{r+1} = ^8C_r (x)^{8-r} \left( -\frac{1}{2x} \right)^r}

Vedclass · Accepted Answer

$-7$

Vedclass · Answer

(C) The general term in the expansion of ${\left( x - \frac{1}{2x} ight)^8}$ is given by ${T_{r+1} = ^8C_r (x)^{8-r} \left( -\frac{1}{2x} ight)^r}$.Simplifying this,we get ${T_{r+1} = ^8C_r \left( -\frac{1}{2} ight)^r x^{8-r} x^{-r} = ^8C_r \left( -\frac{1}{2} ight)^r x^{8-2r}}$.To find the coefficient of ${x^2}$,we set the exponent of $x$ equal to $2$:${8 - 2r = 2}$${2r = 6}$${r = 3}$.Substituting ${r = 3}$ into the expression for the coefficient:Coefficient $= ^8C_3 \left( -\frac{1}{2} ight)^3 = \frac{8 	imes 7 	imes 6}{3 	imes 2 	imes 1} 	imes \left( -\frac{1}{8} ight) = 56 	imes \left( -\frac{1}{8} ight) = -7$.

The coefficient of ${x^2}$ in the expansion of ${\left( x - \frac{1}{2x} \right)^8}$ is

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