Prove usin's Mathematical induction 2^n ≤(n+1)!, for n≥0
1 answer:
Answer:
Step-by-step explanation:
2^0 is less than or equal to 1!, because 1<= 1
if 2^n <= (n+1)!, we wish to show that 2^(n+1) <= (n+2)!, since
(n+2)! = (n+1)! * (n+2), and (n+1)!>= 2^n, then we want to prove that n+2<=2, which is always true for n>=0
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