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Let P (n): 2n < (1 × 2 × 3 × … × n). Then the smallest positive integer for which P(n) is true is A. 1 B. 2 C. 3 D.4 |
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Answer» Given P(n):2n< (1×2×…. ×n) For n=1, 2<2 For n=2, 4<4 For n=3, 6<6 For n=4, 8<24 ∴ the smallest positive integer for which P(n) is true is 4. |
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