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Math Help - [SOLVED] Faulhaber's formula and series

  1. #1
    heather123
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    [SOLVED] Faulhaber's formula and series

    In my assignment, I am asked to find the general formulas for the following series:

    1^2+2^2+3^2+4^2+......+n^2
    1^3+2^3+3^3+4^3+......+n^3
    1^4+2^4+3^4+4^4+......+n^4

    Having worked out the general formulas for the above, i am supposed to note patterns and hence formulate a conjecture for the series

    1^k+2^k+3^k+4^k+......+n^k


    Having worked out the general formulas:


    1^2+2^2+3^2+4^2+......+n^2 = 1/6 n*(n+1)*(2n+1)
    1^3+2^3+3^3+4^3+......+n^3 = 1/4 [n*(n+1)]^2
    1^4+2^4+3^4+4^4+......+n^4 = 1/30 n(n+1)(2n+1)(3n^2+3n-1)

    i have also done some research and found out that the answer to my question is FAULHABER's FORMULA.. but unfortunately I need to be able to note observations and show how to arrive at faulhaber's formula.

    Thanks in advance!!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by heather123 View Post
    In my assignment, I am asked to find the general formulas for the following series:

    1^2+2^2+3^2+4^2+......+n^2
    1^3+2^3+3^3+4^3+......+n^3
    1^4+2^4+3^4+4^4+......+n^4

    Having worked out the general formulas for the above, i am supposed to note patterns and hence formulate a conjecture for the series

    1^k+2^k+3^k+4^k+......+n^k


    Having worked out the general formulas:


    1^2+2^2+3^2+4^2+......+n^2 = 1/6 n*(n+1)*(2n+1)
    1^3+2^3+3^3+4^3+......+n^3 = 1/4 [n*(n+1)]^2
    1^4+2^4+3^4+4^4+......+n^4 = 1/30 n(n+1)(2n+1)(3n^2+3n-1)

    i have also done some research and found out that the answer to my question is FAULHABER's FORMULA.. but unfortunately I need to be able to note observations and show how to arrive at faulhaber's formula.

    Thanks in advance!!
    Try the following conjecture:

    S(n) = 1^k+2^k+3^k+4^k+......+n^k

    is a ploynomial of degree k+1 in n.

    RonL
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