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Which of the following is the base case for 4^n+1 > (n+1)^2 where n = 2?(a) 64 > 9(b) 16 > 2(c) 27 < 91(d) 54 > 8I had been asked this question in a national level competition.Asked question is from Principle of Mathematical Induction topic in section Induction and Recursion of Discrete Mathematics

Answer»

Right choice is (a) 64 > 9

The BEST explanation: Statement By PRINCIPLE of MATHEMATICAL induction, for n=2 the base case of the inequation 4^n+1 > (n+1)^2 should be 64 > 9 and it is TRUE.



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