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Thread: sigma notation

  1. #1
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    sigma notation

    Find $\displaystyle \sum_{r=4}^{n}(lg3^{r})$

    what i did was:

    $\displaystyle =(lg 3) \sum_{r=4}^{n}(r)$
    $\displaystyle =(lg 3)(\frac{1}{2})(n^{2}+7n)$

    i used $\displaystyle \frac{n}{2}[2a+(n-1)d]$ to expand $\displaystyle \sum_{r=4}^{n}(r)$

    but the suggested answer is $\displaystyle (lg 3)(\frac{1}{2})(n-3)(n+4)$........
    where did i go wrong?
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  2. #2
    MHF Contributor red_dog's Avatar
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    The sum has n-3 terms, not n terms.
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  3. #3
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    Quote Originally Posted by red_dog View Post
    The sum has n-3 terms, not n terms.
    then...
    $\displaystyle \frac{n-3}{2}[2(4)+((n-3)+1)]$
    =(1/2)(n-3)(n+6)

    ....i still don't get the (n+4)....@"@
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  4. #4
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by wintersoltice View Post
    then...
    $\displaystyle \frac{n-3}{2}[2(4)+((n-3)+1)]$
    =(1/2)(n-3)(n+6)

    ....i still don't get the (n+4)....@"@
    $\displaystyle \frac{n-3}{2}[2\cdot 4+(n-3-1)]=\frac{(n-3)(n+4)}{2}$
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