October 4th, 2011, 12:35 PM  #1 
Newbie Joined: Oct 2011 Posts: 3 Thanks: 0  Mathematical Induction Proof
I have the equation 2^(k+1) < (k + 3)! for any integer (k) greater or equal to 0. I am supposed to prove it by mathematical induction but I am lost. I already proved the base case, but how am I supposed to prove that the left side will always be less than the right side. So far this is what I have: 2^(k+1) = 2*2*2*2...*2 (k +1) times (k+3)! = 1*2*3*...*(k+1)*(k+2)*(k+3) does anybody know how to prove the top line will always be less than the bottom line. Any Help Would Bw Appreciated 
October 4th, 2011, 01:26 PM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Mathematical Induction Proof
You're using induction, remember? So write 2^(k+1) in terms of 2^k and (k+3)! in terms of (k + 2)! and see if you can prove that if the inequality held for the last pair it will hold for the next.

October 4th, 2011, 01:33 PM  #3 
Newbie Joined: Oct 2011 Posts: 3 Thanks: 0  Re: Mathematical Induction Proof
The form I actually wrote is the (k+1) step. So it is in that form, but I just can't figure out how to prove it logically, even though I know it is true

October 4th, 2011, 01:51 PM  #4  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Mathematical Induction Proof Quote:
Once you do this the problem is actually quite easy.  
October 4th, 2011, 02:26 PM  #5 
Newbie Joined: Oct 2011 Posts: 3 Thanks: 0  Re: Mathematical Induction Proof
The original problem was 2^k < (k+2)!. So then for my inductive step I added k+1 to each side to try to prove that it would work for the next number. So then I got 2^k+1 < (k +1 + 2) which is 2^k+1 < (k+3). I do not understand what forms they are supposed to be in to make the proof. Thanks For The Help 
October 4th, 2011, 05:33 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,762 Thanks: 697  Re: Mathematical Induction Proof
For k=1, 2 < 3!=6 Assume 2^k < (k+2)!, prove 2 ^(k+1) < (k+3)! proof 2^(k+1) = 2x2^k < 2x(k+2)! < (k+3)x(k+2)!=(k+3)! x means multiply 
October 4th, 2011, 09:12 PM  #7  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Mathematical Induction Proof Hello, doleary22! Quote: [color=beige]. . [/color] [color=beige]. . . . . . . . . [/color] [color=beige]. . . . . . [/color]  

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