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Math Help - Really confusing?

  1. #1
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    Really confusing?

    use (1/x) - (1/x+1) to find the following sum.
    (1/1*2) + (1/2*3) + (1/3*4) ..... + (1/99*100)

    show/explain how to get the answer

    the answer is 99/100 if that helps
    Last edited by yoman360; April 27th 2009 at 06:56 PM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by yoman360 View Post
    use (1/x) - (1/x+1) to find the following sum.
    (1/1*2) + (1/2*3) + (1/3*4) ..... + (1/99*100)

    show/explain how to get the answer

    the answer is 99/100 if that helps
    When you expand the sum, something very nice will happen...

    \sum_{x=1}^{99}\left(\frac{1}{x}-\frac{1}{x+1}\right)=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{99}-\frac{1}{100}\right) =1-\frac{1}{100}=\boxed{\frac{99}{100}}

    Does this make sense?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    When you expand the sum, something very nice will happen...

    \sum_{x=1}^{99}{\left(\frac{1}{x}-\frac{1}{x+1}\right)}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{99}-\frac{1}{100}\right) =1-\frac{1}{100}=\boxed{\frac{99}{100}}

    Does this make sense?
    Note that your original series is

    \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots + \frac{1}{99 \times 100} = \sum_{i = 1}^{99}{\frac{1}{x(x + 1)}}


    Also note that \frac{1}{x} - \frac{1}{x + 1} = \frac{1}{x}\times\frac{x + 1}{x + 1} - \frac{1}{x + 1} \times \frac{x}{x}

    = \frac{x + 1}{x(x + 1)} - \frac{x}{x( x + 1)}

     = \frac{x + 1 - x}{x(x + 1)}

     = \frac{1}{x(x + 1)}.


    So \sum_{i = 1}^{99}{\frac{1}{x(x + 1)}} = \sum_{i = 1}^{99}{\left(\frac{1}{x} - \frac{1}{x + 1}\right)}.

    Then refer to Chris' post.
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    When you expand the sum, something very nice will happen...

    \sum_{x=1}^{99}\left(\frac{1}{x}-\frac{1}{x+1}\right)=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{99}-\frac{1}{100}\right) =1-\frac{1}{100}=\boxed{\frac{99}{100}}

    Does this make sense?
    Wait How did you get 1-(1/100)?
    Last edited by yoman360; April 28th 2009 at 04:33 PM.
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  5. #5
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    Quote Originally Posted by yoman360 View Post
    Wait How did you get 1-(1/100)?
    -\frac{1}{2} + \frac{1}{2} = 0

    -\frac{1}{3} + \frac{1}{3} = 0

    -\frac{1}{4} + \frac{1}{4} = 0 etc.


    So everything cancels except 1 - \frac{1}{100}.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    -\frac{1}{2} + \frac{1}{2} = 0

    -\frac{1}{3} + \frac{1}{3} = 0

    -\frac{1}{4} + \frac{1}{4} = 0 etc.


    So everything cancels except 1 - \frac{1}{100}.
    oh thanks man. I don't know why I didn't notice
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