Question #341159

prove that n! > 2n for n a positive integer greater than or qual to 4 what is the base step



1
Expert's answer
2022-05-19T04:26:08-0400

We prove the required result by mathematical induction

The result is true for n = 4

24>16    4!>2424>16\implies 4!>2^4

Let the result be true for n = k. That is

k!>2kk!>2^k Where k4k\ge4

Now we need to prove that the result is also true for n = k + 1. That is

(k+1)!>2k+1(k+1)!>2^{k+1}

By our assumption

k!>2kk!>2^k

    k!(k+1)>2k(k+1)\implies k!(k+1)>2^k(k+1)

    (k+1)!>2k(2)\implies (k+1)!>2^k(2) k+1>2\because k+1>2 replacing k+1k+1 with 2 will not effect the inequality

    (k+1)!>2k+1\implies (k+1)!>2^{k+1}

Hence the result is true for n=k+1n=k+1 . Hence by the principle of mathematical induction the result is true for all n4Z+n\ge4\isin \Z^+



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