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Math Help - Set of values of n for which >10000

  1. #1
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    Set of values of n for which >10000

    Given that \sum^n_{r=1}r(2r-1)=\frac{1}{6}n(n+1)(4n-1)

    Find the set of values of n for which \sum^n_{r=1}(6r^2-3r+2^r)>10000

    \sum^n_{r=1}(6r^2-3r+2^r)=3\sum^n_{r=1}r(2r-1)+\sum^n_{r=1}2^r

    =3[\frac{1}{6}n(n+1)(4n-1)]+\frac{2(1-2^n)}{1-2}

    =\frac{1}{2}n(n+1)(4n-1)-2+2^{n+1}

    How do I solve for n?
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  2. #2
    Grand Panjandrum
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    Re: Set of values of n for which >10000

    Quote Originally Posted by Punch View Post
    Given that \sum^n_{r=1}r(2r-1)=\frac{1}{6}n(n+1)(4n-1)

    Find the set of values of n for which \sum^n_{r=1}(6r^2-3r+2^r)>10000

    \sum^n_{r=1}(6r^2-3r+2^r)=3\sum^n_{r=1}r(2r-1)+\sum^n_{r=1}2^r

    =3[\frac{1}{6}n(n+1)(4n-1)]+\frac{2(1-2^n)}{1-2}

    =\frac{1}{2}n(n+1)(4n-1)-2+2^{n+1}

    How do I solve for n?
    Get your calculator out and try a few values

    You may need to prove that there is but a single positive root to

    \frac{1}{2}n(n+1)(4n-1)-2+2^{n+1}=1000.


    CB
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  3. #3
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    Re: Set of values of n for which >10000

    Quote Originally Posted by CaptainBlack View Post
    Get your calculator out and try a few values

    You may need to prove that there is but a single root to
    \frac{1}{2}n(n+1)(4n-1)-2+2^{n+1}=1000.

    CB
    Does that mean that I have to use a guess and check method?
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  4. #4
    Grand Panjandrum
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    Re: Set of values of n for which >10000

    Quote Originally Posted by Punch View Post
    Does that mean that I have to use a guess and check method?
    That would do it, yes. But don't forget you need to show this has a single positive root.
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  5. #5
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    Re: Set of values of n for which >10000

    Quote Originally Posted by CaptainBlack View Post
    That would do it, yes. But don't forget you need to show this has a single positive root.
    Does that mean I have to show b^2-4ac=0? But how do i go about doing that for an equation with a power
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  6. #6
    Grand Panjandrum
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    Re: Set of values of n for which >10000

    Quote Originally Posted by Punch View Post
    Does that mean I have to show b^2-4ac=0? But how do i go about doing that for an equation with a power
    No, what you need to do is show that:

    f(x)=\frac{1}{2}x(x+1)(4x-1)-2+2^{x+1}-1000

    is monotonic and changes sign between x=0 and say x=1000, then since f(x) is continuous it has a single root and it lies between 0 and 1000.

    CB
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