"In the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n, n \in \mathrm{~N}$, if the ratio of $15^{\text {th }}$ term from the beginning to the $15^{\text {th }}$ term from the end is $\frac{1}{6}$, then the value of ${ }^n \mathrm{C}_3$ is"

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Accepted Answer

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In the expansion of $(a+b)^n$

$15^{\text {th }}$ term from beginning: $T_{15}={ }^n C_{14} a^{n-14} b^{14}$

$15^{\text {th }}$ term from end: $T_{15}^{\prime}={ }^n C_{14} b^{n-14} a^{14}$

$$\begin{aligned} & \therefore \quad \frac{T_{15}}{T_{15}^{\prime}}=\frac{1}{6} \\ & \left(\frac{a}{b}\right)^{n-28}=\frac{1}{6} \\ & \left(6^{\frac{1}{3}}\right)^{n-28}=6^{-1} \\ & \Rightarrow \quad \frac{n-28}{3}=-1 \\ & n=25 \\ & \therefore \quad{ }^{25} C_3=2300 \end{aligned}$$

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In the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n, n \in \mathrm{~N}$, if the ratio of $15^{\text {th }}$ term from the beginning to the $15^{\text {th }}$ term from the end is $\frac{1}{6}$, then the value of ${ }^n \mathrm{C}_3$ is

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In the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n, n \in \mathrm{~N}$, if the ratio of $15^{\text {th }}$ term from the beginning to the $15^{\text {th }}$ term from the end is $\frac{1}{6}$, then the value of ${ }^n \mathrm{C}_3$ is

Solution

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