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Which of the following is the base case for 4^n+1 > (n+1)^2 where n = 2?

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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 > 8

I 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

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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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