The numerically greatest term in the expansio | vedclass

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(C) Given expansion is $(3x - 16y)^{15}$.Substitute $x = \frac{2}{3}$ and $y = \frac{3}{2}$:$3x = 3 	imes \frac{2}{3} = 2$$16y = 16 	imes \frac{3}{2

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$15^{	ext{th}}$ term

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(C) Given expansion is $(3x - 16y)^{15}$.Substitute $x = \frac{2}{3}$ and $y = \frac{3}{2}$:$3x = 3 	imes \frac{2}{3} = 2$$16y = 16 	imes \frac{3}{2} = 24$So,the expression becomes $(2 - 24)^{15} = 2^{15}(1 - 12)^{15}$.Let the expansion be $A(1 + \alpha)^n$,where $A = 2^{15}$,$\alpha = -12$,and $n = 15$.The numerically greatest term $T_{r+1}$ is determined by the condition $r \le \frac{(n+1)|\alpha|}{|\alpha|+1}$.$r \le \frac{(15+1)|-12|}{|-12|+1} = \frac{16 	imes 12}{13} = \frac{192}{13} \approx 14.76$.Since $r$ must be an integer,the greatest value is $r = 14$.Therefore,the $(r+1)^{	ext{th}}$ term,which is the $15^{	ext{th}}$ term,is the numerically greatest term.

The numerically greatest term in the expansion of $(3x - 16y)^{15}$ when $x = \frac{2}{3}$ and $y = \frac{3}{2}$ is

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