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dustbin
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Homework Statement
I am asked to prove:
2n < (n+1)! , where n≥2
The Attempt at a Solution
Base step: set n=2, then test 22 < (2+1)!
22 = 4
(2+1)!= 3! = 3(2)(1) = 6
so 4 < 6 , which is true.
Induction hypothesis is 2k < (k+1)!
Using this, prove 2(k+1) < [(k+1)+1]! = (k+2)!
Attempt to solve:
starting with what I know: 2k < (k+1)!
Multiplying both sides by 2: 2(2k) = 2(k+1) < 2(k+1)!
I know that 2(k+1)! < (k+2)!
since (k+2)! = (k+2)(k+1)! and because k≥2, (k+2) will be greater than 2. Thus, multiplying (k+1)! by 2 on the LHS is less than multiplying (k+1)! by (k+2) on the RHS.
Thus, since 2(k+1) < 2(k+1)! is true, then 2k+1 < [(k+1)+1]!.
P(k+1) follows from P(k), completing the induction step. By mathematical induction, P(n) is true for n≥2.Thanks for any help!
EDIT: fixed a couple of type-o's.
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