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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\left(2^{1 / 3}ight)^{\mathrm{n}-\mathrm{r}}\left(\frac{1}{3^{1 / 3}}ight)^{\mathrm{r}} \ & \mathrm{r}=14 \ & \mathrm{~T}_{15}={ }^{\mathrm{n}} \mathrm{C}_{14}\left(2^{1 / 3}ight)^{\mathrm{n}-14}\left(\frac{1}{3^{1 / 3}}ight)^{14} \end{aligned}$ $\mathrm{T}_{15}^{\prime}=15^{\mathrm{h}}$ term from last is $(\mathrm{n}-13)^{\mathrm{h}}$ term from beginning. $\begin{aligned} & \mathrm{T}_{15}^{\prime}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-14}\left(2^{1 / 3}ight)^{14}\left(\frac{1}{3^{1 / 3}}ight)^{\mathrm{n}-14} \ & \Rightarrow \frac{\mathrm{~T}_{15}}{\mathrm{~T}_{15}^{\prime}}=\frac{{ }^{\mathrm{n}} \mathrm{C}_{14}\left(2^{1 / 3}ight)^{\mathrm{n}-14}\left(\frac{1}{3^{1 / 3}}ight)^{14}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}-14}\left(2^{1 / 3}ight)^{14}\left(\frac{1}{3^{1 / 3}}ight)^{\mathrm{n}-14}}=\frac{1}{6}\end{aligned}$ $\begin{aligned} & =\left(2^{1 / 3}ight)^{n-28}\left(3^{1 / 3}ight)^{n-28}=\frac{1}{6} \ & =6^{\frac{n-28}{3}}=6^{-1}\end{aligned}$ $	ext { So, }{ }^n \mathrm{C}_3={ }^{25} \mathrm{C}_3=2300$

In the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n, \mathrm{n} \in \mathrm{N}$, if the…

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