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Homework Help: First Year Calculus Course Mathematical Induction Problem

  1. Mar 3, 2013 #1
    Prove by mathematical induction:

    (2n)! < (2^(2n))*(n!)^2 , for all n=2,3,4...

    I know that to start you must prove that it is true for n=2,

    (2*2)! = 24 < 64 = (2^4)(2!)^2

    Then you assume that n=k and show tha n=k implies that n=(k+1)

    (2k)! < (2^(2k))*(k!)^2

    ... At this point I am completely lost, I don't know where to go from here to turn it into (k+1)
    Any help would be greatly appreciated.
  2. jcsd
  3. Mar 3, 2013 #2


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    But what factor does each side increase when you go from k to k+1?
  4. Mar 3, 2013 #3
    I'm not too sure... I tried multiplying both sides by 4, because if you do that you would end up getting:

    4(2k)! < 4(2^(2k))*(k!)^2
    4(2k)! < (2^2)*(2^(2k))*(k!)^2
    4(2k)! < (2^(2k+2))*(k!)^2
    4(2k)! < (2^(2(k+1)))*(k!)^2

    But then that still leaves me with (2k) on the left side and (k) on the right.
    Any suggestions? Am I on the right track?
  5. Mar 3, 2013 #4


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    It might be easier to see another way (2(k+1))!=A*(2k)! What's A? And 2^(2(k+1))*(k+1)!^2=B*(2^k*k!^2). What's B?
  6. Mar 3, 2013 #5
    Solving for A I get A = 1 + 1/k, and for B I get B = 4(k+1)^2.
    I don't think that we can multiply the inequality by k, so is there another way to approach this?
  7. Mar 3, 2013 #6


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    B is right. A isn't. (2(k+1))!/(2k)!=(2k+2)(2k+1), isn't it? Just bear with me. The whole point of this exercise is which is larger, A or B?
  8. Mar 3, 2013 #7
    Yes my mistake, A = (2k+2)(2k+1) = (4k^2 + 6k + 2) and B = (4k^2 + 8k + 4), so B is larger.
  9. Mar 3, 2013 #8


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    Yes it is.

    Can you prove it ?
  10. Mar 3, 2013 #9
    Yes! Thank you!

    (2(k+1))! = (2k+2)!
    = (2k+2)(2k+1)!
    = (2k+2)(2k+1)(2k)!
    = (4k^2 + 6k + 2)(2k)!
    < (4k^2 + 8k + 4)(2k)! < (4k^2 + 8k + 4)(2^(2k))(k!)^2 (based on assumption)
    and the rest is algebra!
    Thanks again!
  11. Mar 4, 2013 #10


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    Sure. That's what I was getting at.
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