Proving inequality by mathematical inductionby dustbin Tags: induction, inequality, mathematical, proving 

#1
May912, 04:10 PM

P: 239

1. The problem statement, all variables and given/known data
I am asked to prove: 2^{n} < (n+1)! , where n≥2 3. The attempt at a solution Base step: set n=2, then test 2^{2} < (2+1)! 2^{2} = 4 (2+1)!= 3! = 3(2)(1) = 6 so 4 < 6 , which is true. Induction hypothesis is 2^{k} < (k+1)! Using this, prove 2^{(k+1)} < [(k+1)+1]! = (k+2)! Attempt to solve: starting with what I know: 2^{k} < (k+1)! Multiplying both sides by 2: 2(2^{k}) = 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 2^{k+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 typeo's. 



#2
May912, 05:41 PM

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P: 21,063





#3
May912, 06:54 PM

P: 239

Thank you very much for your response! It was posted as an extra credit problem by my professor. I wanted to make sure my reasoning was correct before posting it on the board, as I've been having a little difficulty grasping what we covered about mathematical induction.




#4
May912, 07:15 PM

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P: 21,063

Proving inequality by mathematical induction
It looks to me like you have the idea of induction proofs.



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