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Prove that (n(top)sigma r=1(bottom))^2-(n-1(top)sigma r=1(bottom))^2 =n^3 general formula for both notation is r

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  1. (sum(r) for r=1..n)^2 - (sum(r) for r=1..n-1)^2 = n^3

    well, we know that

    n
    ∑ r = n(n+1)/2
    r=1
    so, square that and we have

    n^2(n+1)^2/4

    Sum to n-1 instead of n, and we have

    (n-1)^2 n^2/4

    Subtract to get

    n^2/4 ((n+1)^2 - (n-1)^2)
    = n^2/4 (4n)
    = n^3

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    posted by Steve

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