Question

if the seventh term from the beginning and end in the binomial expansion of  \[\left( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \right)^n\] are equal, find n.

 

 

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,

\[T_7 = T_{n - 5} \]

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

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

\[ \Rightarrow \left( 6 \right)^\frac{n}{3} - 2 = 6^2 \]

\[ \Rightarrow \frac{n}{3} - 2 = 2\]

\[ \Rightarrow n = 12\]

Hence, the value of n is 12.