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Math Help - Inequalities

  1. #1
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    Inequalities

    Can anyone help me confirm if I've got this one correct? Particularly the part at A, where I've defined (k + 2)! = (k + 1)!(k + 2).

    Many thanks.

    Q.
    (n + 1)! > 2n, for n \in \mathbb{N}

    Attempt: Step 1: For n = 1...
    (1 + 1)! = 2! = 2 & 21 = 2
    Since 2 > 2, the statement is true for n = 1.

    Step2: Assume the statement is true for n = k, i.e. assume (k + 1)! > 2k.
    We must now show that the statement is true for n = k + 1,
    i.e. ((k + 1) + 1)! > 2k+1 if (k + 2)! > 2k+1
    A: (k + 2)! = (k + 1)!(k + 2) > (k + 1)2k...((k + 1)! > 2k...assumed)
    If (k + 1)2k > 2k+1, then (k + 2)! > 2k+1
    if (k + 1)2k > 2k+1
    if (k + 1)2k - 2k+1 > 0
    if 2k[(k + 1) - 2] > 0...which is true for k > 1
    => (k + 2)! > 2k+1
    Therefore, the statement is true for n = k + 1, if true for n = k. Thus, the statement is true for all n > 2, n \in \mathbb{N}
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  2. #2
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    Re: Inequalities

    Quote Originally Posted by GrigOrig99 View Post
    Can anyone help me confirm if I've got this one correct? Particularly the part at A, where I've defined (k + 2)! = (k + 1)!(k + 2).
    Many thanks. Q. (n + 1)! > 2n, for n \in \mathbb{N}
    Attempt: Step 1: For n = 1...
    (1 + 1)! = 2! = 2 & 21 = 2
    Since 2 > 2, the statement is true for n = 1.
    Step2: Assume the statement is true for n = k, i.e. assume (k + 1)! > 2k.
    We must now show that the statement is true for n = k + 1,
    i.e. ((k + 1) + 1)! > 2k+1 if (k + 2)! > 2k+1
    A: (k + 2)! = (k + 1)!(k + 2) > (k + 1)2k...((k + 1)! > 2k...assumed)
    If (k + 1)2k > 2k+1, then (k + 2)! > 2k+1
    if (k + 1)2k > 2k+1
    if (k + 1)2k - 2k+1 > 0
    if 2k[(k + 1) - 2] > 0...which is true for k > 1
    => (k + 2)! > 2k+1
    Therefore, the statement is true for n = k + 1, if true for n = k. Thus, the statement is true for all n > 2, n \in \mathbb{N}
    You tend to make too much of this.
    Part of step 2 should note that assuming K\ge 2 we have (K+1)!\ge 2^K. So you kn ow that K+2>2.
    Look at [(K+2)!=(K+2)(K+1)!\ge(2)(2^K)=2^{K+1}
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  3. #3
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    Re: Inequalities

    Ok, so A should be rewritten as: (k + 2)! = (k + 1)!(k + 2) > (k + 2)2k...?
    Last edited by GrigOrig99; September 6th 2012 at 07:32 AM.
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  4. #4
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    Re: Inequalities

    Quote Originally Posted by GrigOrig99 View Post
    Ok, so A should be rewritten as: (k + 2)! = (k + 1)!(k + 2) > (k + 2)2k...?
    Yes.
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  5. #5
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    Re: Inequalities

    Cont. from there:
    If (k + 2)2k > 2k+1, then (k + 2)! > 2k+1
    (k + 2)2k > 2k+1
    if (k + 2)2k - 2k+1 > 0
    if 2k[(k + 2) - 2k] > 0...true when k > 1
    => (k + 2)! > 2k+1
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