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Math Help - elem.num.theo - help w/proof

  1. #1
    Member cassiopeia1289's Avatar
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    elem.num.theo - help w/proof

    So - I know all the steps, I just need help with a gap.

    Prove n! > n^2      \exists     n \geq4
    Ok, proof by Induction:

    Since 4! = 24 which is >(4)^2 = 16, n!>n^2 for n=4.
    Let k be a positive integer [math\geq4[/tex] and suppose k!>k^2.
    Then, k!(k+1) > k^2(k+1)
    (k+1)! > k^2(k+1) = k^3+k^2

    Here's where I'm stuck. I know k\geq4 but how can I show that (k+1)^2 < k^3+k^2?

    From then, I can show (k+1)! > k^3+k^2 > (k+1)^2
    Therefore, (k+1)! > (k+1)^2, so n! > n^2        for        n \geq4.

    Please help me out with this way of solving the proof, I'm sure there are other ways.

    any help: please!!!??
    Last edited by cassiopeia1289; September 25th 2008 at 07:28 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cassiopeia1289 View Post
    So - I know all the steps, I just need help with a gap.

    Prove n! > n^2      \exists     n \geq4
    Ok, proof by Induction:

    Since 4! = 24 which is >(4)^2 = 16, n!>n^2 for n=4.
    Let k be a positive integer [math\geq4[/tex] and suppose k!>k^2.
    Then, k!(k+1) > k^2(k+1)
    (k+1)! > k^2(k+1) = k^3+k^2

    Here's where I'm stuck. I know k\geq4 but how can I show that (k+1)^2 < k^3+k^2?

    From then, I can show (k+1)! > k^3+k^2 > (k+1)^2
    Therefore, (k+1)! > (k+1)^2, so n! > n^2        for        n \geq4.

    Please help me out with this way of solving the proof, I'm sure there are other ways.

    any help: please!!!??
    you will have that k^3 + k^2 > (k + 1)^2 if you can show k^3 > 2k + 1 for k \ge 4
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  3. #3
    Member cassiopeia1289's Avatar
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    so how do I go from (k+1)!>k^3+k^2 to k^3>2k+1 - where did the 2k+1 come from?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cassiopeia1289 View Post
    so how do I go from (k+1)!>k^3+k^2 to k^3>2k+1
    well, if i told you that, i'd be doing the problem for you, wouldn't i?

    - where did the 2k+1 come from?
    we want to show that

    k^3 + k^2 > (k + 1)^2 = k^2 + {\color{red}2k + 1}

    thus we need to show that k^3 > 2k + 1
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