Principle Of Mathematical Induction Question 61
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}}} $
Correct Answer: B
Show Answer
Answer:
Solution:
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
