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 ${\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)}^n$, ${\left[\left(n+1\right)-7+1\right]}^{\mathrm{th}}$ 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,
$$\begin{gathered} T_7=T_{n-5} \\ {\Rightarrow }^n{C}_6\times 2^{\frac{n}{3}-2}\times \frac{1}{3^2}{=}^n{C}_6\times 2^2\times \frac{1}{3^{{\displaystyle \frac{n}{3}-2}}} \\ \Rightarrow \frac{2^{\frac{n}{3}-2}}{2^2}=\frac{3^2}{3^{{\displaystyle \frac{n}{3}}-2}} \\ \Rightarrow {\left(6\right)}^{\frac{n}{3}-2}=6^2 \\ \Rightarrow \frac{n}{3}-2=2 \\ \Rightarrow n=12 \end{gathered}$$

Hence, the value of n is 12.