The numerically greatest term in the expansio | vedclass

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(B) Given expansion is $(2x - 3y)^{11}$. Substituting $x = \frac{1}{3}$ and $y = \frac{1}{2}$,we get $(2(\frac{1}{3}) - 3(\frac{1}{2}))^{11} = (\frac{

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${ }^{11}C_3 \left(\frac{3}{2}\right)^5$

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(B) Given expansion is $(2x - 3y)^{11}$. Substituting $x = \frac{1}{3}$ and $y = \frac{1}{2}$,we get $(2(\frac{1}{3}) - 3(\frac{1}{2}))^{11} = (\frac{2}{3} - \frac{3}{2})^{11} = (\frac{2}{3})^{11} (1 - \frac{3/2}{2/3})^{11} = (\frac{2}{3})^{11} (1 - \frac{9}{4})^{11}$.For the expansion $(1 + a)^n$,the numerically greatest term $T_{r+1}$ is determined by $r = \lfloor \frac{(n+1)|a|}{|a|+1} floor$.Here $n = 11$ and $a = -\frac{9}{4}$,so $|a| = \frac{9}{4} = 2.25$.$r = \lfloor \frac{(11+1)(2.25)}{2.25+1} floor = \lfloor \frac{12 	imes 2.25}{3.25} floor = \lfloor \frac{27}{3.25} floor = \lfloor 8.307 floor = 8$.Thus,the $9^{	ext{th}}$ term $(T_9)$ is the greatest term.$T_9 = { }^{11}C_8 (\frac{2}{3})^{11-8} (-\frac{3}{2})^8 = { }^{11}C_3 (\frac{2}{3})^3 (\frac{3}{2})^8 = { }^{11}C_3 (\frac{2}{3})^3 (\frac{3}{2})^3 (\frac{3}{2})^5 = { }^{11}C_3 (1)^3 (\frac{3}{2})^5 = { }^{11}C_3 (\frac{3}{2})^5$.

The numerically greatest term in the expansion of $(2x - 3y)^{11}$ when $x = \frac{1}{3}$ and $y = \frac{1}{2}$ is:

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