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Prove by induction

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove by induction that n2>(4n+7) for all integers n[tex]\geq[/tex]6.
    [Hint: somewhere you may use the fact that (2k+1)[tex]\geq[/tex]13 > 4.]

    2. Relevant equations

    3. The attempt at a solution

    n2> (4n+7)

    (B) when n =6
    62> (4*6+7)
    36>31 true

    (I) given n=k then k2> (4k+7)

    let n=k+1= k2+2k+1

    (k2+ 2k+1) > 4k+7 +2k+1

    4k+7+2k+1> 4k+7+4

    6k+8 > 4(k+1)+7

    Is that looks like okay? Please help. Thank you.
  2. jcsd
  3. Nov 9, 2008 #2


    Staff: Mentor

    It's hard to follow the logic in what you wrote.

    For the step you're trying to prove, you should have something like this:
    (k + 1)^2 > ... > 4(k + 1) + 7
    In other words, your inequality starts off with (k + 1)^2 and ends with a smaller quantity, 4(k + 1) + 7.
  4. Nov 10, 2008 #3
    Okay. Thank you.
  5. Nov 10, 2008 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    Surely this is not what you meant to write! you meant, I think, that "let n= k+1, then n2= (k+1)2= k2+ 2k+ 1

    [/quote](k2+ 2k+1) > 4k+7 +2k+1

    4k+7+2k+1> 4k+7+4 [/quote]
    It would be useful to point out that because k> 5, 2k+1> 11> 4

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