Mathematics - Binomial Theorem Question with Solution | TestHub

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