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Strong Induction - Integrals

  1. Jan 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove, for r [tex]\geq[/tex] 1,

    Sr(n) = [tex]\sum[/tex] kr = (nr+1)/(r+1) + Pr(n)

    where Pr(n) is some polynomial with respect to n of degree r.
    (Hint: Prove using strong induction on r. After you show the base case is true, assume the statement istrue for all r = 1,2, ... ,R-1 and then show the statement is true for r = R.)


    2. Relevant equations
    (r + 1) [tex]\sum[/tex] kr = n(r+1) + Pr(n) - [tex]\sum[/tex] [c2kr-1 + c3kr-2 + ... + crk + 1], where c# is some constant.


    3. The attempt at a solution
    I tried to prove the base case (r=1), but I can't seem to complete it.
    Using the equation given,
    Sr(n) = [tex]\sum[/tex] kr = (nr+1)/(r+1)
    -> the left side goes to: n2
    And plugging in r=1, the right side goes to n2/2 + Pr(n)

    Am I missing something? The two sides aren't equal at all, and I'm at a lost.
     
  2. jcsd
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