In the expansion of (√[3]2+ 1 √[3]3 )^n, n ∈ N, if the ratio of 15^ at term from the beginning to the 15^ th term from the end is 1 6, then the value…

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