Question

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

• A. $5$
• B. $7$
• C. $13$
• D. $Noneofthese$

Answer 1

The correct option is C $13$

As we know that in an expansion ${(a+b)}^n$, the general term is given as-

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

In the given expansion-

${(2^{1/3}+\frac{1}{3^{1/3}})}^n$

$a=2^{1/3},b=\frac{1}{3^{1/3}},n=?$

Therefore,

$\Rightarrow \frac{{}^n{C}_5{\left(2^{1/3}\right)}^5{\left(\frac{1}{3^{1/3}}\right)}^{n-5}}{{}^n{C}_{n-5}{\left(2^{1/3}\right)}^{n-5}{\left(\frac{1}{3^{1/3}}\right)}^5}=\frac{1}{6}\left(\text{Given}\right)$

$\Rightarrow \frac{1}{{\left(2^{1/3}\right)}^{n-10}{\left(3^{1/3}\right)}^{n-10}}=\frac{1}{6}$

$\Rightarrow \frac{1}{{\left(6\right)}^{\frac{n-10}{3}}}=\frac{1}{{\left(6\right)}^1}$

$\Rightarrow \frac{n-10}{3}=1$

$\Rightarrow n-10=3$

$\Rightarrow n=10+3=13$