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Math Help - Sum of Series

  1. #1
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    Sum of Series

    Q) Find the sum of the series:
    a).
    1/(2^1/2 + 1^1/2) + 1/(2^1/2 + 3^1/2) + .......+ 1/(120^1/2 +121^1/2)

    b).
    1/(1 x 5) + 1/(5 x 9) + 1/(9 x 13) + .......+ 1/(221 x 225).


    Any help would be appreciated.

    Thanks
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  2. #2
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    Quote Originally Posted by a69356 View Post
    Q) Find the sum of the series:
    a).
    1/(2^1/2 + 1^1/2) + 1/(2^1/2 + 3^1/2) + .......+ 1/(120^1/2 +121^1/2)

    b).
    1/(1 x 5) + 1/(5 x 9) + 1/(9 x 13) + .......+ 1/(221 x 225).
    These are both telescoping series. For a), the n'th term is \frac1{\sqrt{n+1}+\sqrt n} = \sqrt{n+1}-\sqrt n. For b), the n'th term is \frac1{(4n-3)(4n+1)}, which you can split into partial fractions.
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  3. #3
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    Thanks Opalg,

    I found the sum of the first series it comes out to be 10.

    But for the second one I found that there are 56 terms in the series and the series is neither A.P nor G.P.

    How do we proceed further to find the sum of it.

    Thanks,
    Ashish
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  4. #4
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    \frac1{(4n-3)(4n+1)} = \frac{1/4}{4n-3} - \frac{1/4}{4n+1}. So

    \begin{aligned}\frac1{1\times5} +  \frac1{5\times9}\, +  &\,\frac1{9\times13} + \ldots +  \frac1{221\times225}\\ &= \frac14\biggl(\Bigl(\frac11-\frac15\Bigr) + \Bigl(\frac15-\frac19\Bigr) + \Bigl(\frac19-\frac1{13}\Bigr) + \ldots + \Bigl(\frac1{221}-\frac1{225}\Bigr)\biggr).\end{aligned}
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