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. |
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