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Math Help - summation

  1. #1
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    summation

    Find expression in terms of n for the sum of this series , to the number of terms stated , simplifynig the answer as far as possible .

    2^3+4^3+6^3+8^3+... to n terms .

    My work :

    \sum^{n}_{r=1}r^3-\sum^{n}_{r=1} (2r-1)^3

    =\frac{1}{4}n^2(n+1)^2-\sum^{n}_{r=1}(8r^3-12r^2+6r-1)

    =\frac{1}{4}n^2(n+1)^2-8[\frac{1}{4}n^2(n+1)^2]+12(\frac{1}{6}n(n+1)(2n+1))-6(\frac{1}{2}n(n+1))+n

    I am not really sure how to put it into its simplest form from here .

    i can only see that n is common .
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  2. #2
    MHF Contributor red_dog's Avatar
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    (2\cdot 1)^3+(2\cdot 2)^3+(2\cdot 3)^3+\ldots+(2n)^3=

    =\sum_{k=1}^n(2k)^3=8\sum_{k=1}^nk^3=8\cdot\frac{n  ^2(n+1)^2}{4}=2n^2(n+1)^2
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