Binomial Theorem And Its Simple Applications Question 37
Question: If the ratio of the 7th term from the beginning to the 7th term from the end in $ {{( \sqrt[3]{2}+\frac{1}{\sqrt[3]{3}} )}^{n}} $ is $ \frac{1}{6} $ them n equals to
Options:
A) 10
B) 9
C) 8
D) 12
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Given $ \frac{T_7}{{T_{n-7+2}}}=\frac{1}{6}\Rightarrow \frac{T_7}{{T_{n-s}}}=\frac{1}{6} $
$ \Rightarrow \frac{^{n}C_6{{( \sqrt[3]{2} )}^{n-6}}{{( \frac{1}{\sqrt[3]{3}} )}^{6}}}{^{n}{C_{n-6}}{{( \sqrt[3]{2} )}^{6}}{{( \frac{1}{\sqrt[3]{3}} )}^{n-6}}}=\frac{1}{6} $
$ \Rightarrow {2^{\frac{n-12}{3}}}.,{3^{\frac{n-12}{3}}}=\frac{1}{6}\Rightarrow {6^{\frac{n-12}{3}}}={6^{-1}} $
$ \therefore \frac{n-12}{3}=-1\Rightarrow n=9 $
BETA
