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Thread: sum

  1. #1
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    sum

    hi

    Am i correct to think

    $\displaystyle \Sigma k^3 = \Sigma k^2 \times k = \Sigma k^2 \times \Sigma k, \forall k$?

    help please
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  2. #2
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    Quote Originally Posted by 1cookie View Post
    Am i correct to think
    $\displaystyle \Sigma k^3 = \Sigma k^2 \times k = \Sigma k^2 \times \Sigma k, \forall k$?
    $\displaystyle \sum\limits_{k = 1}^3 {k^3 } = 36\;,\;\sum\limits_{k = 1}^3 {k^2 } = 14\;\& \;\sum\limits_{k = 1}^3 k = 6$

    Now what do you think?
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  3. #3
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    Quote Originally Posted by Plato View Post
    $\displaystyle \sum\limits_{k = 1}^3 {k^3 } = 36\;,\;\sum\limits_{k = 1}^3 {k^2 } = 14\;\& \;\sum\limits_{k = 1}^3 k = 6$

    Now what do you think?

    ...hmm, clearly $\displaystyle 14 \times 6 \neq 36$


    i see your point Plato!
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