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Question

Let P(n) : 2n<(1×2×3×....×n). Then the smallest positive integer for which P(n) is true is


A

1

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B

2

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C

3

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D

4

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Solution

The correct option is D

4


(d) 4

Since,

P(1) : 2<1 is false

P(2) : 22<1×2 is false

P(3) : 23<1×2×3 is false

But P(4) : 24<1×2×3×4 is true.


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