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

Now,
$$\begin{gathered} T_7{=}^n{C}_6{\left(\sqrt[3]{2}\right)}^{n-6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^6{=}^n{C}_6\times 2^{\frac{n}{3}-2}\times \frac{1}{3^2} \\ \mathrm{And}, \\ {T}_{n-5}{=}^n{C}_{n-6}{\left(\sqrt[3]{2}\right)}^6{\left(\frac{1}{\sqrt[3]{3}}\right)}^{n-6}{=}^n{C}_6\times 2^2\times \frac{1}{3^{{\displaystyle \frac{n}{3}}-2}} \end{gathered}$$

It is given that,
$$\begin{gathered} \frac{T_7}{T_{n-5}}=\frac{1}{6} \\ \Rightarrow \frac{{}^n{C}_6\times 2^{\frac{n}{3}-2}\times \frac{1}{3^2}}{{}^n{C}_6\times 2^2\times \frac{1}{3^{{\displaystyle \frac{n}{3}-2}}}}=\frac{1}{6} \\ \Rightarrow 2^{\frac{n}{3}-2-2}\times 3^{\frac{n}{3}-2-2}=\frac{1}{6} \\ \Rightarrow {\left(\frac{1}{6}\right)}^{4-\frac{n}{3}}=\frac{1}{6} \\ \Rightarrow 4-\frac{n}{3}=1 \\ \Rightarrow n=9 \end{gathered}$$

Hence, the value of n is 9.