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Math Help - Use the properties of summation to evaluate the sum

  1. #1
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    Use the properties of summation to evaluate the sum


    Could someone tell me how to solve this?
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  2. #2
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    Re: Use the properties of summation to evaluate the sum

    I would use the formulas:

    \sum_{k=1}^n(1)=n

    \sum_{k=1}^n(k)=\frac{n(n+1)}{2}

    \sum_{k=1}^n(k^2)=\frac{n(n+1)(2n+1)}{6}

    \sum_{k=1}^n(k^3)=\frac{n^2(n+1)^2}{4}

    and use the property \sum_{k=1}^n(a\cdot f(k)\pm b\cdot g(k))=a\cdot\sum_{k=1}^n(f(k))\pm b\cdot\sum_{k=1}^n(g(k)) to rewrite the sum first, then use the above formulas.
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    Re: Use the properties of summation to evaluate the sum

    Is this correct?
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    Re: Use the properties of summation to evaluate the sum

    Quote Originally Posted by calculus123 View Post
    Is this correct?
    NO! It is not.

    It should be \sum\limits_{k = 1}^{10} {k^3 }  - 2\sum\limits_{k = 1}^{10} {k^2 }  + 5\sum\limits_{k = 1}^{10} k  - \sum\limits_{k = 1}^{10} {(1)}

    See reply #2 to finish.
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  5. #5
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    Re: Use the properties of summation to evaluate the sum

    I have spent all day today trying to solve it...I can't. Can someone please solve it and show me to do it?
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Use the properties of summation to evaluate the sum

    \sum_{k=1}^{10}(k^3)-2\sum_{k=1}^{10}(k^2)+5\sum_{k = 1}^{10}(k)-\sum_{k = 1}^{10}(1)=

    \frac{10^2\cdot11^2}{4}-2\cdot\frac{10\cdot11\cdot21}{6}+5\cdot\frac{10 \cdot11}{2}-10=

    3025-770+275-10=2520
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  7. #7
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    Re: Use the properties of summation to evaluate the sum

    Quote Originally Posted by calculus123 View Post
    I have spent all day today trying to solve it...I can't. Can someone please solve it and show me to do it?
    You really need to have a live tutor.

    Quote Originally Posted by MarkFL2 View Post
    I would use the formulas:
    \sum_{k=1}^n(1)=n
    \sum_{k=1}^n(k)=\frac{n(n+1)}{2}
    \sum_{k=1}^n(k^2)=\frac{n(n+1)(2n+1)}{6}
    \sum_{k=1}^n(k^3)=\frac{n^2(n+1)^2}{4}
    Use the above:

    \sum\limits_{k = 1}^{10} {k^3 }  - 2\sum\limits_{k = 1}^{10} {k^2 }  + 5\sum\limits_{k = 1}^{10} k  - \sum\limits_{k = 1}^{10} {(1)}

    Now plug in:
     \frac{10^2(10+1)^2}{4}-2\left(\frac{10(10+1)(2\cdot 10+1)}{6}\right)+5\left( \frac{10(10+1)}{2}\right)- 10
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