Question

In the expansion of ${(3\sqrt{2}+\frac{1}{3\sqrt{3}})}^n$ the ratio of $7^{th}$ term from the beginning and from the end is $6^{667}$, then $n$=.....

• A. $667$
• B. $2001$
• C. $6$
• D. $2013$

Answer 1

The correct option is D $2013$

giben binomial expression ${(3\sqrt{2}+\frac{1}{3\sqrt{3}})}^n$

So, according to question

$\frac{{}^n{C}_6(3\sqrt{2}{)}^{n-6}{\left(\frac{1}{3\sqrt{3}}\right)}^6}{{}^n{C}_6{\left(\frac{1}{3\sqrt{3}}\right)}^{n-6}(3\sqrt{2}{)}^6}$

$\Rightarrow \left(3\sqrt{2}{)}^{n-12}\right(3\sqrt{3}{)}^{n-12}=(6{)}^{667}$

$\Rightarrow (3\sqrt{2}3\sqrt{3}{)}^{n-12}=(6{)}^{667}$

$\Rightarrow (6{)}^{\frac{n-12}{3}}=(6{)}^{667}$

$\because$ base is same $\frac{n-12}{3}=667\Rightarrow n=2013$

So, option $\left(d\right)$ is correct