The numerically greatest term in the binomial | vedclass

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(D) Given the expression $(2a - 3b)^{19}$.Substitute $a = \frac{1}{4}$ and $b = \frac{2}{3}$:$2a = 2(\frac{1}{4}) = \frac{1}{2}$$3b = 3(\frac{2}{3}) =

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$^{19}C_3 \cdot 2^{13}$

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(D) Given the expression $(2a - 3b)^{19}$.Substitute $a = \frac{1}{4}$ and $b = \frac{2}{3}$:$2a = 2(\frac{1}{4}) = \frac{1}{2}$$3b = 3(\frac{2}{3}) = 2$So,the expression becomes $(\frac{1}{2} - 2)^{19} = (\frac{1}{2})^{19}(1 - 4)^{19}$.Let $T_r$ be the $r^{th}$ term in the expansion of $(x+y)^n$. The numerically greatest term $T_{r+1}$ satisfies $r \le \frac{(n+1)|y|}{|x|+|y|}$.Here $n=19$,$x=1$,$y=-4$.$r \le \frac{(19+1)|-4|}{|1|+|-4|} = \frac{20 	imes 4}{5} = 16$.Since $r=16$ is an integer,both $T_{16}$ and $T_{17}$ are the numerically greatest terms.Calculating $T_{17} = ^{19}C_{16} (\frac{1}{2})^{19-16} (-2)^{16} = ^{19}C_3 (\frac{1}{2})^3 (2^{16}) = ^{19}C_3 \cdot 2^{-3} \cdot 2^{16} = ^{19}C_3 \cdot 2^{13}$.

The numerically greatest term in the binomial expansion of $(2a - 3b)^{19}$ when $a = \frac{1}{4}$ and $b = \frac{2}{3}$ is

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