Binomial Theorem And Its Simple Applications Question 73
Question: If $ \frac{4^{n}}{n+1}<\frac{(2n)!}{{{(n!)}^{2}}}, $ then P(n) is true for
Options:
A) $ n\ge 1 $
B) $ n>0 $
C) $ n<0 $
D) $ n\ge 2 $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] Let $ P(n):\frac{4^{n}}{n+1}<\frac{(2n)!}{{{(n!)}^{2}}} $
For $ n=2;P(2):\frac{4^{2}}{2+1}<\frac{4!}{{{(2)}^{2}}}\Rightarrow \frac{16}{3}<\frac{24}{4} $
Which is true. Let for $ n=m\ge 2,P(m) $ is true. i.e., $ \frac{4^{m}}{m+1}<\frac{(2m)!}{{{(m!)}^{2}}} $
Now, $ \frac{{4^{m+1}}}{m+2} $
$ =\frac{4^{m}}{m+1}.\frac{4(m+1)}{m+2}<\frac{(2m)!}{{{(m!)}^{2}}}.\frac{4(m+1)}{(m+2)} $
$ =\frac{(2m)!(2m+1)(2m+2)4(m+1){{(m+1)}^{2}}}{(2m+1)(2m+2){{(m!)}^{2}}{{(m+1)}^{2}}(m+2)} $
$ =\frac{[2(m+1)]!}{{{[(m+1)!]}^{2}}}.\frac{2{{(m+1)}^{2}}}{(2m+1)(m+2)}<\frac{[2(m+1)]!}{{{[(m+1)!]}^{2}}} $
Hence, for $ n\ge 2, $ P(n) is true.
BETA
