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Math Help - Summation Help

  1. #1
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    Summation Help

    Hi all,

    Just starting off learning summations, and I'm looking to find if there is an easier way to solve this problem.

    <br /> <br />
\sum (\frac{1}{k^2} - \frac{1}{(k-1)^2})<br /> <br />

    The sum is k = 4, up to 24, sorry didn't know how to put it in for latex.

    I know what it's asking, but surely there is a faster way than manually substituting in values and adding.

    Any help is appreciated.
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  2. #2
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    Krizalid's Avatar
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    That's actually a telescoping sum. Do you know what a telescoping sum is?
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  3. #3
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    Nope, any chance you can point me in the right direction?

    I'm guessing it has to do with the fact it's 1/16 - 1/9, then 1/25 - 1/16 etc etc, with the numbers shifting, but I'm not sure of the best way to deal with it.
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  4. #4
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  5. #5
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    Quote Originally Posted by Peleus View Post
    Hi all,

    Just starting off learning summations, and I'm looking to find if there is an easier way to solve this problem.

    <br /> <br />
\sum (\frac{1}{k^2} - \frac{1}{(k-1)^2})<br /> <br />

    The sum is k = 4, up to 24, sorry didn't know how to put it in for latex.

    I know what it's asking, but surely there is a faster way than manually substituting in values and adding.

    Any help is appreciated.
    \sum_{k=4}^{24} \frac{1}{k^2} = \frac{1}{4^2} + \frac{1}{5^2} + ... + \frac{1}{24^2}

    \sum_{k=4}^{24} -\frac{1}{(k-1)^2} = -\frac{1}{3^2} - \frac{1}{4^2} - ... - \frac{1}{23^2}<br />

    sum these two series and what two terms remain?
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  6. #6
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    Thanks guys,

    So essentially I'm left with

    <br />
\frac{1}{24^2} - \frac{1}{3^2}<br />

    For the answer.


    Appreciate it folks.
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