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Math Help - find the value for this expression?

  1. #1
    Member grgrsanjay's Avatar
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    find the value for this expression?

    find the value of \sum_{k=1}^{99}\frac{1}{K\sqrt{K+1} + (K+1)\sqrt{K}})
    Last edited by grgrsanjay; August 29th 2010 at 03:18 AM.
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    Quote Originally Posted by grgrsanjay View Post
    find the value of \sum_{k=1}^{99}\frac{1}{K\sqrt{K+1} + (K+1)\sqrt{K}})
    Sketch solution: Rationalise the fraction by multiplying top and bottom by (k+1)\sqrt k - k\sqrt{k+1}, and you will find that you have a telescoping sum.

    Spoiler:
    The answer is 9/10.
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  3. #3
    Member grgrsanjay's Avatar
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    Hello, grgrsanjay!

    Opalg gave you an excellent game plan . . . Did you follow it?



    \displaystyle \text{Find the value of: }\;\sum_{k=1}^{99}\,\frac{1}{(k\!+\!1)\sqrt{k} + k\sqrt{k\!+\!1}}

    Rationalize the denominator . . . multiply top and bottom by the conjugate:

    . . \displaystyle \frac{1}{(k\!+\!1)\sqrt{k} + k\sqrt{k\!+\!1}} \cdot\frac{(k\!+\!1)\sqrt{k} - k\sqrt{k\!+\!1}}{(k\!+\!1)\sqrt{k} - k\sqrt{k\!+\!1}} \;\;=\;\; \frac{(k\!+\!1)\sqrt{k} - k\sqrt{k\!+\!1}}{(k\!+\!1)^2k - k^2(k\!+\!1)} \;\;=\;\;\frac{(k\!+\!1)\sqrt{k} - k\sqrt{k\!+\!1}}{k(k\!+\!1)}


    . . . . \displaystyle =\;\;\frac{(k\!+\!1)\sqrt{k}}{k(k\!+\!1)} - \frac{k\sqrt{k\!+\!1}}{k(k\!+\!1)} \;\;=\;\;\frac{\sqrt{k}}{k} - \frac{\sqrt{k\!+\!1}}{k\!+\!1} \;\;=\;\;\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k\!+\!1}}



    The sum becomes: . \displaystyle \sum^{99}_{k=1}\left(\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k\!+\!1}}\right)

    . . . . . . . . . . . . \displaystyle =\;\;\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) + \hdots + \left(\frac{1}{\sqrt{99}} - \frac{1}{\sqrt{100}}\right)



    All terms cancel out . . . except the first and the last.

    . . The sum is: . \displaystyle \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{100}} \;\;=\;\;1 - \frac{1}{10} \;\;=\;\;\boxed{\frac{9}{10}}

    thanks you sorobon and thanks opalg
    Last edited by grgrsanjay; September 1st 2010 at 05:17 PM.
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