Question

If the ratio of the $7^{th}$ term from the beginning to the $7^{th}$ term from the end in the expansion of
${(2^{1/3}+\frac{1}{3^{1/3}})}^n$ is $\frac{1}{6}$, find $n$.

• A. $5$
• B. $9$
• C. $7$
• D. $2$

Answer 1

The correct option is B $9$

The $7^{th}$ from begining of the expansion ${(2^{\frac{1}{3}}+\frac{1}{3^{\frac{1}{3}}})}^n$

$T_{6+1}$ $=$ ${}^n{C}_6$.$2^{\frac{n-6}{3}}$.$\frac{1}{3^{\frac{6}{3}}}$ $=$ ${}^n{C}_6$.$2^{\frac{n-6}{3}}$.$\frac{1}{9}$

To find $7^{th}$ term from end we just find $7^{th}$ term of the expansion ${(\frac{1}{3^{\frac{1}{3}}}+2^{\frac{1}{3}})}^n$ from beginning.

$T{"}_{6+1}$ $=$ ${}^n{C}_6$.$\frac{1}{3^{\frac{n-6}{3}}}$.$2^{\frac{6}{3}}$ $=$ ${}^n{C}_6$. $\frac{1}{3^{\frac{n-6}{3}}}$.$4$

$\Rightarrow$$\frac{T_7}{T{"}_7}$ $=$ $\frac{1}{6}$

$\Rightarrow$$\frac{1}{6}$ $=$ $\frac{{}^n{C}_6{.2}^{\frac{n-6}{3}}.\frac{1}{9}}{{}^n{C}_6.\frac{1}{3^{\frac{n-6}{3}}}.4}$

$\Rightarrow$$\frac{1}{6}$ $=$ $\frac{6^{\frac{n-6}{3}}}{36}$

$\Rightarrow$$6^1$ $=$ $6^{\frac{n-6}{3}}$

On equating the power, we get

$\Rightarrow$ $1=$ $\frac{n-6}{3}$

So, $n=9$