Question

If the ratio of the $7th$ term from the beginning to the $7th$ term from the end in the expansion of ${\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)}^n$ is $\frac{1}{6}$, then $n=$

• A. 6
• B. 8
• C. 9
• D. 12

Answer 1

Answer: (C)

Since, the general term in the expansion of ${\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)}^n$ is
$T_{r+1}={}^n{C}_r{\left(2^{1/3}\right)}^{n-r}{\left(\frac{1}{3^{1/3}}\right)}^r$
Now, the $7th$ term from beginning $={}^n{C}_6{2}^{\frac{n-6}{3}}{3}^{-2}$
and the $7th$ term from end $={}^n{C}_6{\left(\frac{1}{3^{1/3}}\right)}^{n-6}{\left(2^{1/3}\right)}^6$
$\because$ It is given that,
$\frac{{}^n{C}_6{2}^{\frac{n-6}{3}}{3}^{-2}}{{}^n{C}_6{3}^{\frac{6-n}{3}}{2}^2}=\frac{1}{6}$
$\Rightarrow \frac{1}{\frac{6-n}{3}+2}=\frac{1}{6}$
$\Rightarrow \frac{6-n}{3}+2=1$
$\Rightarrow \frac{6-n}{3}=-1$
$\Rightarrow 6-n=-3$
$\Rightarrow n=9$