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Math Help - Sigma Notation

  1. #1
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    Sigma Notation

    \Sigma [i=1, 15] i(i-1)^2
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by yeloc View Post
    \Sigma [i=1, 15] i(i-1)^2
    \sum_{n=1}^{15}n(n-1)^2=\sum_{n=0}^{15}(n+1)n^2=\sum_{n=1}^{15}n^3+n^  2

    Now apply your known formulas
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  3. #3
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    Can you explain how you went from the first step to the second step?
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  4. #4
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    Quote Originally Posted by Mathstud28 View Post
    \sum_{n=1}^{15}n(n-1)^2=\sum_{n=0}^{15}(n+1)n^2=\sum_{n=1}^{15}n^3+n^  2

    Now apply your known formulas
    Quote Originally Posted by yeloc View Post
    Can you explain how you went from the first step to the second step?
    I can see why you'd be a bit puzzled ..... here's the trick:

    Substitute n - 1 = m \Rightarrow n = m+1 into \sum_{n=1}^{15}n(n-1)^2. Then you get:

    \sum_{m=0}^{{\color{red}14}} (m+1) m^2 = \sum_{m={\color{red}0}}^{14} (m^3+m^2).

    Note the corrections (in red) to some typos.
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  5. #5
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    Please forgive me, but I am still not understanding why you substitute  n=m+1
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  6. #6
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    Quote Originally Posted by yeloc View Post
    Please forgive me, but I am still not understanding why you substitute  n=m+1
    Because that's what Mathstud28 did to get the result that you asked about. If you don't like that, just expand the original summand:

    \sum_{i=1}^{15} i (i - 1)^2 = \sum_{i=1}^{15} (i^3 - 2i^2 + i) = \sum_{i=1}^{15} i^3 - 2 \sum_{i=1}^{15} i^2 + \sum_{i=1}^{15}i.

    Now use the usual formulae on each term.
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  7. #7
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    Ohh ok...thanks!
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    I can see why you'd be a bit puzzled ..... here's the trick:

    Substitute n - 1 = m \Rightarrow n = m+1 into \sum_{n=1}^{15}n(n-1)^2. Then you get:

    \sum_{m=0}^{{\color{red}14}} (m+1) m^2 = \sum_{m={\color{red}0}}^{14} (m^3+m^2).

    Note the corrections (in red) to some typos.
    Thank you for catching that typo...after doing so many infinite sreies you just forget to lower the upper index.
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