Question

Find n in the binomial \[\left( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \right)^n\] , if the ratio of 7^th term from the beginning to the 7^th term from the end is  \[\frac{1}{6}\]

 

 

Answer 1

In the binomail expansion of  \[\left( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \right)^n\] \[\left[ \left( n + 1 \right) - 7 + 1 \right]^{th}\]  i.e., (n − 5)^th term from the beginning is the 7^th term from the end.
Now,

\[T_7 =^n C_6 \left( \sqrt[3]{2} \right)^{n - 6} \left( \frac{1}{\sqrt[3]{3}} \right)^6 =^n C_6 \times 2^\frac{n}{3} - 2 \times \frac{1}{3^2}\]

\[\text{ And, }\]

\[ T_{n - 5} =^n C_{n - 6} \left( \sqrt[3]{2} \right)^6 \left( \frac{1}{\sqrt[3]{3}} \right)^{n - 6} =^n C_6 \times 2^2 \times \frac{1}{3^\frac{n}{3} - 2}\]

It is given that,

\[\frac{T_7}{T_{n - 5}} = \frac{1}{6}\]

\[ \Rightarrow \frac{{}^n C_6 \times 2^\frac{n}{3} - 2 \times \frac{1}{3^2}}{{}^n C_6 \times 2^2 \times \frac{1}{3^\frac{n}{3} - 2}} = \frac{1}{6}\]

\[ \Rightarrow 2^\frac{n}{3} - 2 - 2 \times 3^\frac{n}{3} - 2 - 2 = \frac{1}{6}\]

\[ \Rightarrow \left( \frac{1}{6} \right)^{4 - \frac{n}{3}} = \frac{1}{6}\]

\[ \Rightarrow 4 - \frac{n}{3} = 1\]

\[ \Rightarrow n = 9\]

Hence, the value of n is 9.