A) \[n>2\]
B) \[n\in N\]
C) \[n>3\]
D) None of these
Correct Answer: A
Solution :
[a] Let \[P(n)\equiv n!>{{2}^{n-1}};P(3)\equiv 6>4\] Let \[P(k)\equiv k!>{{2}^{k-1}}\] is true. \[\therefore P(k+1)=(k+1)!=(k+1)k!>(k+1){{2}^{k-1}}\] \[>{{2}^{k}}\,(as\,\,k+1>2)\]
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