Question

Find $n$ in the binomial

$(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}}{)}^n$ if the ratio of $7^{th}$ term from the beginning to the $7^{th}$ term from the end is $\frac{1}{6}$

Answer 1

The general term of the expansion $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

Given expression is

${(\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}})}^n$

$T_{r+1}{=}^n{C}_r\left(\sqrt[3]{32}{)}^{n-r}{\left(\frac{1}{\sqrt[3]{33}}\right)}^r{=}^n{C}_r\right(2{)}^{\frac{n-r}{3}}(3{)}^{-\frac{r}{3}}$

So, the 7th term from the beginning is

$T_7=T_{6+1}{=}^n{C}_6\left(2{)}^{\frac{n-6}{3}}\right(3{)}^{\frac{-6}{3}}$

$\Rightarrow T_7{=}^n{C}_6\left(2{)}^{\frac{n-6}{3}}\right(3{)}^{-2}$

The $p^{th}$ term from the end is same as $(n-p+2{)}^{th}$ term from the beginning, so the 7th term from the end is same as $(n-5{)}^{th}$ term from the beginning, which is

$T_{n-5}=T_{n-6+1}{=}^n{C}_{n-6}\left(2{)}^{\left(\frac{n-(n-6)}{3}\right)}\right(3{)}^{\left(\frac{-(n-6)}{3}\right)}$

$\Rightarrow T_{n-5}{=}^n{C}_{n-6}\left(2{)}^2\right(3{)}^{-\frac{(n-6)}{3}}$

Given, the ratio of $7^{th}$ term from the beginning to the $7^{th}$ term from the end is $\frac{1}{6}$ .

$\Rightarrow \frac{{}^n{C}_6\left(2{)}^{\frac{n-6}{3}}\right(3{)}^{-2}}{{}^n{C}_{n-6}\left(2{)}^2\right(3{)}^{-}\frac{(n-6)}{3}}=\frac{1}{6}$

$\Rightarrow \frac{\left(2{)}^{\frac{n-6}{3}}\right(3{)}^{-2}}{\left(2{)}^2\right(3{)}^{-}\frac{(n-6)}{3}})=\frac{1}{6}$

$\left[{\because }^n{C}_6{=}^n{C}_{n-6}\right]$

$\Rightarrow (2{)}^{\frac{(n-6)}{3}}-2(3{)}^{-}2+\frac{(n-6)}{3}=\frac{1}{6}$

$\Rightarrow \left(2{)}^{\frac{(n-12)}{3}}\right(3{)}^{\frac{(n-12)}{3}}=\frac{1}{6}$

$\Rightarrow (2\times 3{)}^{\frac{(n-12)}{3}}=\frac{1}{6}\Rightarrow (6{)}^{\frac{(n-12)}{3}}=6^{-1}$

$\Rightarrow \frac{n-12}{3}=-1\Rightarrow n-12=-3$

$\therefore n=9$

Hence, the value of $n$ is $9$.