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Math Help - Series and Sequences, Sigma notation.

  1. #1
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    Series and Sequences, Sigma notation.

    Given that r^2 +1 = (r+2)(r+1) - 3(r+1) + 2

    Show that Sigma r=1 r=n (r^2 + 1)r! = n(n+1)!

    Help please ><
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  2. #2
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    Re: Series and Sequences, Sigma notation.

    I understand that you're new here, welcome to MHF. Next time, you may want to wrap [tex] tags, so your question will look like this:

    Given that r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2

    Show that \sum_{r=1}^{r=n} (r^2 + 1)r! = n(n+1)!.


    Anyway, replace r^2 + 1 with (r+2)(r+1) - 3(r+1) + 2 so that our summation becomes


    \sum_{r=1}^{r=n} ( (r+2)(r+1) - 3(r+1) + 2) r!


    =\sum_{r=1}^{r=n} (r+2)! - 3(r+1)! + 2r!


    =(3! + ... + (n+2)!) - 3(2! + ... + (n+1)!) + 2(1! + ... + n!)


    Note that if 3 \le i \le n, all of the i! terms cancel out, leaving


    (n+1)! + (n+2)! - 3(2! + (n+1)!) + 2(1! + 2!)


    =(n+2)! - 2(n+1)!


    =(n+2)(n+1)! - 2(n+1)!, this factors to n(n+1)!.
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  3. #3
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    Re: Series and Sequences, Sigma notation.

    OMG Thank You
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