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Thread: 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 $\displaystyle n! > n^2 \exists n \geq4$
    Ok, proof by Induction:

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

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

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

    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; Sep 25th 2008 at 06:28 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 $\displaystyle n! > n^2 \exists n \geq4$
    Ok, proof by Induction:

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

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

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

    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 $\displaystyle k^3 + k^2 > (k + 1)^2$ if you can show $\displaystyle k^3 > 2k + 1$ for $\displaystyle k \ge 4$
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  3. #3
    Member cassiopeia1289's Avatar
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    so how do I go from $\displaystyle (k+1)!>k^3+k^2$ to $\displaystyle k^3>2k+1$ - 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 $\displaystyle (k+1)!>k^3+k^2$ to $\displaystyle k^3>2k+1$
    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

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

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