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Math Help - Miscellaneous series

  1. #1
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    Unhappy Miscellaneous series

    1^2+(1^2+2^2)+(1^2+2^2+3^2)+............. n terms
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by matsci0000 View Post
    1^2+(1^2+2^2)+(1^2+2^2+3^2)+............. n terms
    This is the same as \sum_{k=1}^n\sum_{j=1}^kj^2

    So \sum_{k=1}^n\sum_{j=1}^kj^2=\sum_{k=1}^n\frac{k(k+  1)(2k+1)}{6}=\sum_{k=1}^n\frac{k^3}{3}+\frac{k^2}{  2}+\frac{k}{6} =\frac{1}{3}\left[\frac{n(n+1)}{2}\right]^2+\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{12}=\tfra  c{1}{12}n(n+2)(n+1)^2
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