Question

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

• A. $6$
• B. $7$
• C. $8$
• D. $9$

Answer 1

Answer: (D)

$9$

Seventh term from beginning:

$t_7={{}^{n}C}_r{\left(2^{\frac{1}{3}}\right)}^{n-6}{\left(3^{-\frac{1}{3}}\right)}^6$

Seventh term from end is $(n-5{)}^{th}$ term from beginning

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

$t_7:t_{n-5}=1:6$

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

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

$\Rightarrow \frac{n-12}{3}=-1\Rightarrow n=9$