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 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
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Re: Mathematical Induction Proof

Quote:
 Originally Posted by doleary22 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
You wrote down what (k+3)! is, but not what it is in terms of (k+2)!. Ditto for 2^(k+1) and 2^k.

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
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Re: Mathematical Induction Proof

Hello, doleary22!

Quote:
 $\text{Prove by induction: }\:2^n \:<\:(n+2)!$

$\text{Verify }S(1):\;\;2^1 \:<\:3!\; \text{ . . . True!}$

$\text{Assume }S(k):\;\;2^k \:<\:(k+2)!$

$\text{We want to prove }S(k+1):\;\;2^{k+1}\:<\: (k+3)!$

$\text{Multiply }S(k)\text{ by 2: }\;2\,\cdot\,2^k \:<\:2\,\cdot\,(k+2)!$

[color=beige]. . [/color]$\text{and we have: }\:2^{k+1} \:<\:2(k+2)!\;\;[1]$

$\text{For }k\,\ge\,0,\,\text{ we have: }\:2\;<\; k+3$

$\text{Multiply by }(k+2)!\;\;\;\;2\,\cdot\,(k+2)!\;<\;(k+2)!\,\cdot\ ,(k+3)$

[color=beige]. . . . . . . . . [/color]$\text{That is: }\;\;\;\;\;2(k+2)! \;<\;(k+3)!\;\;[2]$

$\text{From [1] and [2], we have: }\;2^{k+1} \;<\;2(k+2)! \;<\;(k+3)!$

[color=beige]. . . . . . [/color]$\text{Therefore: }\;2^{k+1} \;<\;(k+3)!$

$\text{W\!e have proved }S(k+1).\;\text{ The inductive proof is complete.}$

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