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October 4th, 2011, 01:35 PM   #1
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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
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October 4th, 2011, 02:26 PM   #2
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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.
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October 4th, 2011, 02:33 PM   #3
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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
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October 4th, 2011, 02:51 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.
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October 4th, 2011, 03:26 PM   #5
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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
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October 4th, 2011, 06:33 PM   #6
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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
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October 4th, 2011, 10:12 PM   #7
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Re: Mathematical Induction Proof

Hello, doleary22!

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