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  1. #1
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    algebra problem


    " alt="\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
    " />

    is there a quick way to find out wha the sum of the series would be
    if u reolace the n with 1, then 2, then 3 up to.... 99?

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  2. #2
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    Your equation is not shown.
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  3. #3
    Super Member craig's Avatar
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    Quote Originally Posted by the undertaker View Post
    \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}
    Your equation did not show due all the colour tags you had in your post.
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  4. #4
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    \sum_{i=1}^{99} \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}

    maybe?
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  5. #5
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by the undertaker View Post
    \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}

    is there a quick way to find out wha the sum of the series would be
    if u reolace the n with 1, then 2, then 3 up to.... 99?
    If you write out the sum, you have

    \sum_{n\,=\,1}^{99}\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\ =\ \frac1{\sqrt1}-\frac1{\sqrt2}\,+\,\frac1{\sqrt2}-\frac1{\sqrt3}\,+\,\frac1{\sqrt3}-\frac1{\sqrt4}\,+\cdots+\,\frac1{\sqrt{99}}-\frac1{\sqrt{100}}

    you can see that all the terms cancel except the first and the last. Hence the sum is 1-\frac1{10}=\frac9{10}.
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