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Sum of the first n powers

  1. Jul 24, 2007 #1
    1. The problem statement, all variables and given/known data
    I encountered this formula in a proof:

    Let [tex]S_k(n)[/tex] be the sum of the first n positive integers raised to the power of k:

    e.g. [tex]S_3(5) = 1^3 + 2^3+3^3+4^3+5^3[/tex]

    How would one show that

    [tex]S_2(n) = (n^3/3) + (n^2/2) + an[/tex]

    and that

    [tex]S_4(n) = (n^5/5) + (n^4/2) +bn^3 + cn^2 +dn [/tex]

    where a,b,c,d are constants.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 24, 2007 #2
    attempt? there needs to be at least some work b4 help is given.
     
  4. Jul 24, 2007 #3
    3. The attempt at a solution

    n^5/5 is the integral of n^4 and n^3/3 is the integral of n^2
    However, that is only true for the first term.
     
  5. Jul 24, 2007 #4
    When you're given a formula and asked to prove it, it's often helpful to try induction.
     
  6. Jul 25, 2007 #5

    Dick

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    You are guessing e.g. for S4 that it is a fifth degree polynomial. Then do something in the spirit of induction. What is S4(k)-S4(k-1)? Once you answer that you can solve for the coefficients.
     
  7. Jul 25, 2007 #6
    S4(k)-S4(k-1) = k^4
    So, I should set that equal to a(k)^5 + b(k)^4 + c(k)^3 +d(k)^2 + e(k) + f - [a(k-1)^5 + b(k-1)^4 + c(k-1)^3 +d(k-1)^2 + e(k-1) + f] ?
    I am not sure how you solve that or how you would do induction on that.
     
  8. Jul 25, 2007 #7

    Dick

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    Expand everything out and move everything to one side. Now you've got a polynomial that must equal zero for all values of k. So the coefficients of all powers of k must be zero.
     
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