Question #160238

By using mathematical induction prove that (n+1)! > 2^(n+1) for n, where n is a positive integer greater than or equal to 4


1
Expert's answer
2021-02-04T06:52:19-0500

Prove (n+1)!>2(n+1)(n+1)! > 2^{(n+1)} :


Base case: n=4n = 4

(4+1)!=5!=120>2(4+1)=25=32(4+1)! = 5! = 120 > 2^{(4+1)} = 2^5 = 32 -> true


suppose for some n: n!>2nn! > 2^n , let's prove that (n+1)!>2(n+1)(n+1)! > 2^{(n+1)}

(n+1)!=n!(n+1)(n+1)! = n! * (n+1)

2(n+1)=22n2^{(n+1)} = 2 * 2^n

using assumption that n!>2nn! > 2^n it is obvious that n!(n+1)>22nn! * (n+1) > 2 * 2^n , since n+1>2n+1 > 2 for n4n \ge 4 .


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