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Induction help

  1. Oct 12, 2009 #1
    Hi,

    I need some help with mathematical induction

    The question is as follows:

    prove that [tex]\sum_{i=0}^n (i-3) \geq \frac{n^2}{4}[/tex]

    I have shown that it holds for the base step where n=12
    [tex]\frac{144}{4} = 36[/tex]
    and the sum of all the i's up to 12 [tex]-3,-2,-1,0,+1,+2,+3,+4,+5,+6,+7,+8,+9 = 39[/tex]

    [tex]39\geq36[/tex]

    Now for the inductive step:
    [tex] \sum_{i=0}^{n+1} [/tex]
    and i know this can be rewritten, but im not sure how
    either it's [tex]\sum_{i=0}^{n} (i-3)(i-3)~~ \text{or}~~\sum_{i=0}^{n} (i-3)+(i-3)[/tex]

    How do i proceed from here?
     
  2. jcsd
  3. Oct 12, 2009 #2

    lanedance

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    i think you need
    [tex]\sum_{i=0}^{n+1} (i-3) = ((n+1)-3) + \sum_{i=0}^{n} (i-3)[/tex]

    then assuming the proposition is true for n leads to
    [tex]\sum_{i=0}^{n+1} (i-3) = ((n+1)-3) + \sum_{i=0}^{n} (i-3) \geq ((n+1)-3) + \frac{n^2}{4}[/tex]
     
  4. Oct 12, 2009 #3

    HallsofIvy

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    The last term, that you want to take out of the sum, has i= n+1 so i- 3= n+1-3= n- 2.

    [tex]\sum_{i=0}^{n+1}(i- 3)= \sum_{i=0}^n (i- 3)+ (n+1- 3)= \sum_{i=0}^n (i-3)+ n- 2[/tex]
     
  5. Oct 12, 2009 #4
    Okay,
    Then you have to add (n+1) to the other side of the equation as well?
    [tex]\sum_{i=0}^n (i-3)+ n- 2 = \frac{(n+1)^2}{4} + (n+1)[/tex]
     
  6. Oct 12, 2009 #5

    HallsofIvy

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    ?? No, add n- 2 to other side as well!!
     
  7. Oct 14, 2009 #6
    Haha, im so bad at this, it's almost funny :biggrin:
     
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