The ratio of the 5^{th} term from the beginni | vedclass

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(C) Let the expansion be $(a+b)^n$ where $a = 2^{1/3}$,$b = \frac{1}{2(3)^{1/3}}$,and $n = 10$.The $r^{th}$ term from the beginning is $T_r = {}^{n}C_

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$4(36)^{1/3} : 1$

Vedclass · Answer

(C) Let the expansion be $(a+b)^n$ where $a = 2^{1/3}$,$b = \frac{1}{2(3)^{1/3}}$,and $n = 10$.The $r^{th}$ term from the beginning is $T_r = {}^{n}C_{r-1} a^{n-r+1} b^{r-1}$.The $5^{th}$ term from the beginning is $T_5 = {}^{10}C_4 (2^{1/3})^6 (\frac{1}{2(3)^{1/3}})^4 = {}^{10}C_4 (2^2) \frac{1}{2^4 (3)^{4/3}} = {}^{10}C_4 \frac{1}{2^2 (3)^{4/3}}$.The $5^{th}$ term from the end is the $(10-5+2) = 7^{th}$ term from the beginning.$T_7 = {}^{10}C_6 (2^{1/3})^4 (\frac{1}{2(3)^{1/3}})^6 = {}^{10}C_4 (2^{4/3}) \frac{1}{2^6 (3)^2} = {}^{10}C_4 \frac{1}{2^{14/3} (3)^2}$.The ratio is $\frac{T_5}{T_7} = \frac{{}^{10}C_4 \frac{1}{2^2 (3)^{4/3}}}{{}^{10}C_4 \frac{1}{2^{14/3} (3)^2}} = \frac{2^{14/3}}{2^2} \cdot \frac{3^2}{3^{4/3}} = 2^{8/3} \cdot 3^{2/3} = (2^4 \cdot 3^2)^{1/3} = (16 \cdot 9)^{1/3} = (144)^{1/3} = (4 \cdot 36)^{1/3} = 4(36)^{1/3}$.Thus,the ratio is $4(36)^{1/3} : 1$.

The ratio of the $5^{th}$ term from the beginning to the $5^{th}$ term from the end in the binomial expansion of $\left( 2^{1/3} + \frac{1}{2(3)^{1/3}} \right)^{10}$ is

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