If the ratio of the coefficients of the third | vedclass

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(D) The general term in the expansion of $(x - \frac{1}{2x})^n$ is given by $T_{r+1} = ^nC_r (x)^{n-r} (-\frac{1}{2x})^r = ^nC_r (x)^{n-2r} (-\frac{1}

Vedclass · Accepted Answer

$-10$

Vedclass · Answer

(D) The general term in the expansion of $(x - \frac{1}{2x})^n$ is given by $T_{r+1} = ^nC_r (x)^{n-r} (-\frac{1}{2x})^r = ^nC_r (x)^{n-2r} (-\frac{1}{2})^r$.The third term $(T_3)$ corresponds to $r=2$: $T_3 = ^nC_2 (x)^{n-4} (-\frac{1}{2})^2 = ^nC_2 (\frac{1}{4}) x^{n-4}$.The coefficient of the third term is $C_3 = ^nC_2 	imes \frac{1}{4} = \frac{n(n-1)}{2} 	imes \frac{1}{4} = \frac{n(n-1)}{8}$.The fourth term $(T_4)$ corresponds to $r=3$: $T_4 = ^nC_3 (x)^{n-6} (-\frac{1}{2})^3 = ^nC_3 (-\frac{1}{8}) x^{n-6}$.The coefficient of the fourth term is $C_4 = ^nC_3 	imes (-\frac{1}{8}) = \frac{n(n-1)(n-2)}{6} 	imes (-\frac{1}{8}) = -\frac{n(n-1)(n-2)}{48}$.Given the ratio of coefficients is $1:2$,so $\frac{C_3}{C_4} = \frac{1}{2}$:$\frac{n(n-1)/8}{-n(n-1)(n-2)/48} = \frac{1}{2}$$\frac{n(n-1)}{8} 	imes \frac{-48}{n(n-1)(n-2)} = \frac{1}{2}$$\frac{-6}{n-2} = \frac{1}{2}$$n-2 = -12$$n = -10$.

If the ratio of the coefficients of the third and fourth terms in the expansion of $(x - \frac{1}{2x})^n$ is $1 : 2$,then the value of $n$ is:

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