Question

If the ratio of the $7\,th$ term from the beginning to the $7\, th$ 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 $7\, th$ term from beginning $={ }^{n} C_{6} 2^{\frac{n-6}{3}} 3^{-2}$
and the $7 \,th$ 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 $
$\Rightarrow n=9$