If the ratio of the $$7^{th}$$ term from the beginning to the $$7^{th}$$ term from the end in $${\left( {\root 3 \of 2 + \frac{1}{{\root 3 \of 3 }}} \right)^n}$$   is $$\frac{1}{6}$$ them $$n$$ equals to

Solution :
Given, $$\frac{{{T_7}}}{{{T_{n - 7 + 2}}}} = \frac{1}{6}$$
$$\eqalign{ & \Rightarrow \frac{{^n{C_6}{{\left( {\root 3 \of 2 } \right)}^{n - 6}}{{\left( {\frac{1}{{\root 3 \of 3 }}} \right)}^6}}}{{^n{C_{n - 6}}{{\left( {\root 3 \of 2 } \right)}^6}{{\left( {\frac{1}{{\root 3 \of 3 }}} \right)}^{n - 6}}}} = \frac{1}{6} \cr & \Rightarrow {2^{\frac{{n - 12}}{3}}} \cdot {3^{\frac{{n - 12}}{3}}} = \frac{1}{6} \cr & \Rightarrow {6^{\frac{{n - 12}}{3}}} = {6^{ - 1}} \cr & \therefore \frac{{n - 12}}{3} = - 1 \cr & \Rightarrow n = 9 \cr} $$