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Math Help - Discrete sigma

  1. #1
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    Discrete sigma

    Hi

    Consider the following similarities between the sums:

    1 = 1
    2 + 3 + 4 = 1 + 8
    5 + 6 + 7 + 8 + 9 = 8 + 27
    10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64

    Do you see the pattern? Find a general formula, with sigma and Exponentiations, which summarizes this pattern.

    I need some guiding...
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  2. #2
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    Quote Originally Posted by iHeji View Post
    Hi

    Consider the following similarities between the sums:

    1 = 1
    2 + 3 + 4 = 1 + 8
    5 + 6 + 7 + 8 + 9 = 8 + 27
    10 + 11 + 12 + 13 + 14 + 15 + 16 = 27 + 64

    Do you see the pattern? Find a general formula, with sigma and Exponentiations, which summarizes this pattern.

    I need some guiding...
    Perhaps this will help, perhaps it will not.

    The first equation (N = 1) is given as
    \displaystyle \sum_{n=1}^{1+0}n = (N - 1)^3 + N^3

    The second (N = 2) is
    \displaystyle \sum_{n=2}^{2+2}n = (N - 1)^3 + N^3

    The third (N = 3) is
    \displaystyle \sum_{n=5}^{5+4}n = (N - 1)^3 + N^3

    etc.

    Notice that the lowest number for n is given by the series 1, 2, 5, 10, 17, etc. This is the recursive function
    f(N) = f(N - 1) + (2N - 3),~f(1) = 1

    where N is the equation number. So we have:
    The first equation (N = 1) is given as
    \displaystyle \sum_{n=f(1)}^{f(1)+0}n = (N - 1)^3 + N^3

    The second (N = 2) is
    \displaystyle \sum_{n=f(2)}^{f(2)+2}n = (N - 1)^3 + N^3

    The third (N = 3) is
    \displaystyle \sum_{n=f(3)}^{f(3)+4}n = (N - 1)^3 + N^3

    etc.

    Now notice that the upper term of the summation is of the form f(N) + g(N) where g(n) = 2(N - 1). So the general Nth equation is given by

    \displaystyle \sum_{n = f(N)}^{f(N) + 2(N - 1)}n = (N - 1)^3 + N^3

    This is probably a more complicated method than is needed, but it's the one I came up with that was the most demonstrative. Anyway, now solve f(N) as a function of N (rather than by recursion) and I'd guess the rest would be a simple induction proof.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Go figure. That F(N) function is a simple quadratic. I should have seen that coming.

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    I just finished this. It's takes a while to get an explicit form for f(N), but it is do-able this way. Perhaps one of the other helpers will have a shortcut.

    -Dan
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