In the expansion of left(sqrt[3]{2}+frac{1}{s | vedclass

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(C) The general term of the expansion is $T_{r+1} = {}^n C_r (2^{1/3})^{n-r} (3^{-1/3})^r$.The $15^{	ext{th}}$ term from the beginning is $T_{15} = {

Vedclass · Accepted Answer

$2300$

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(C) The general term of the expansion is $T_{r+1} = {}^n C_r (2^{1/3})^{n-r} (3^{-1/3})^r$.The $15^{	ext{th}}$ term from the beginning is $T_{15} = {}^n C_{14} (2^{1/3})^{n-14} (3^{-1/3})^{14}$.The $15^{	ext{th}}$ term from the end is the $(n-15+1)^{	ext{th}} = (n-14)^{	ext{th}}$ term from the beginning,which is $T'_{15} = T_{n-14+1} = {}^n C_{n-14} (2^{1/3})^{14} (3^{-1/3})^{n-14}$.Given the ratio $\frac{T_{15}}{T'_{15}} = \frac{1}{6}$,and noting ${}^n C_{14} = {}^n C_{n-14}$:$\frac{(2^{1/3})^{n-14} (3^{-1/3})^{14}}{(2^{1/3})^{14} (3^{-1/3})^{n-14}} = \frac{1}{6}$$(2^{1/3})^{n-28} (3^{1/3})^{n-28} = 6^{-1}$$(6^{1/3})^{n-28} = 6^{-1}$$\frac{n-28}{3} = -1$ $\Rightarrow n-28 = -3$ $\Rightarrow n = 25$.Finally,${}^{25} C_3 = \frac{25 	imes 24 	imes 23}{3 	imes 2 	imes 1} = 25 	imes 4 	imes 23 = 2300$.

In the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n, n \in N$,if the ratio of the $15^{\text{th}}$ term from the beginning to the $15^{\text{th}}$ term from the end is $\frac{1}{6}$,then the value of ${}^n C_3$ is:

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