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Thread: Triangular differences

  1. #1
    Member rtblue's Avatar
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    Triangular differences

    Hey everyone,

    I apologize in advance for (potentially) posting this in the wrong forum; not sure exactly which branch of math my problem classifies as. Let me introduce the problem. Let's consider a sequence of squares:

    0, 1, 4, 9, 16, ....

    Let's take the difference between these:

    1, 3, 5, 7,

    And again:

    2, 2, 2, 2...

    Let's try it with cubes:

    1, 8, 27, 64

    7, 19, 37

    12, 18

    6

    You can verify it for yourself, but it turns out the pattern is that given the 2nd difference of a sequences of squares is 2! and the 3rd difference of a sequence of cubes is 3! and the 4th difference of a sequence of n^4 terms is 4!. I'm trying to prove that for a sequence of consecutive integers in the form n^k, (n+1)^k, (n+2)^k, etc., that the kth difference will be k!.

    It's pretty clear that binomial expansion and pascal's triangle pop up here, but I'm unsure of how exactly to rigorously prove this. Maybe induction is the right avenue? Any advice would be appreciated!
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  2. #2
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    skeeter's Avatar
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    Re: Triangular differences

    have a look at the link ... happy researching!

    method-of-differences
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  3. #3
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    Re: Triangular differences

    I would think that strong induction might well work.

    The key expression will be $\displaystyle (m+1)^n - m^n = \sum_{k=0}^{n-1}{n \choose k}m^k$ or it might be easier using $\displaystyle m^n - (m-1)^n = \sum_{k=0}^{n-1}(-1)^{n-k}{n \choose k}m^k$
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  4. #4
    Member rtblue's Avatar
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    Re: Triangular differences

    Thanks!
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