Binomial Theorem And Its Simple Applications Question 133
Question: $ \frac{(2n)!}{2^{2n}{{(n!)}^{2}}}is\le $
Options:
A) $ \frac{1}{3n+1} $
B) $ \frac{1}{{{(3n+1)}^{1/2}}} $
C) $ \frac{1}{{{(3n+1)}^{2}}} $
D) $ \frac{1}{{{(3n+1)}^{1/2}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] By setting $ n=1, $ we get $ \frac{2n!}{2^{2n}.{{(n!)}^{2}}} $ as $ \frac{2}{2^{2}.{{(1)}^{2}}}=\frac{2}{4}=\frac{1}{2}. $
Second alternative gives $ \frac{1}{2} $ . Upon setting $ n=2,\frac{2n!}{2^{2n}.{{(n)}^{2}}} $ becomes $ \frac{3}{8} $ , second alternative Becomes $ \frac{1}{\sqrt{7}} $ which is greater than $ \frac{3}{8}. $
BETA
