JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Question Bank Self Evaluation Test - Principle of Mathematical Induction

If \[\frac{{{4}^{n}}}{n+1}<\frac{(2n)!}{{{(n!)}^{2}}},\] then P(n) is true for

A) \[n\ge 1\]

B) \[n>0\]

C) \[n<0\]

D) \[n\ge 2\]

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.