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Math Help - algebra problem

  1. #1
    Junior Member
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    algebra problem

    1/sqrt(n) - 1/sqrt(n+1)

    is there a quick way to find out the sum of the series where n=1 + n=2 + n=3, all the way up to...99?


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  2. #2
    Super Member
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    Couldn't you have replied to your own thread at http://www.mathhelpforum.com/math-he...-real-one.html ?

    \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
    for n = 1, 2, 3,... 99

    See what happens when you plug in the numbers:

    \left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right) + \left(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\right) + ... + \left(\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\right)

    Everything cancels except the first and last terms!

    \frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}

    \frac{1}{1}-\frac{1}{10}

    \frac{9}{10}


    01
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  3. #3
    MHF Contributor Amer's Avatar
    Joined
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    Jordan
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    1,093
    you mean

    \sum_{n=1}^{99} \frac {1}{\sqrt n} -\frac {1}{\sqrt{n+1}}

    lats expand some

    \frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt4}....\frac{1}{\sqrt{n+1}}

    as you can see the terms delete each other finally you get just

    1-\frac{1}{\sqrt{n+1}}

    now substitute 99 instead of n see what you will get ...
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