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Math Help - Problem related to sequence and series

  1. #1
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    Problem related to sequence and series

    Q- Find the sum of the series?

    1/(2^1/2 + 1^1/2) + 1/(2^1/2 + 3^1/2) + ......... + 1/(120^1/2 + 121^1/2)

    Any help would be greatly appreciated.

    Thanks,
    Ashish
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Write
    1 = sqrt(2)^2 - sqrt(1)^2

    1 = sqrt(3)^2 - sqrt(2)^2

    1 = sqrt(4)^2 - sqrt(3)^2

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    Remember that

    sqrt(a)^2 - sqrt(b)^2 ={sqrt(a) - sqrt(b)} x {sqrt(a) + sqrt(b)}


    Use this formula for each term many things will be cancelled .. ..
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  3. #3
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    Quote Originally Posted by a69356 View Post
    Q- Find the sum of the series?

    1/(2^1/2 + 1^1/2) + 1/(2^1/2 + 3^1/2) + ......... + 1/(120^1/2 + 121^1/2)

    Any help would be greatly appreciated.

    Thanks,
    Ashish
    Note that \frac{1}{{\sqrt {k + 1}  + \sqrt k }} = \sqrt {k + 1}  - \sqrt k .

    You series is then \sum\limits_{k = 1}^{120} {\frac{1}{{\sqrt {k + 1}  + \sqrt k }}}  = \sum\limits_{k = 1}^{120} {\left( {\sqrt {k + 1}  - \sqrt k } \right)}

    Write out several terms to see what is going to happen.
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