JEE Main 2025 — Binomial Theorem Question with Solution

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