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Math Help - Telescopic sums

  1. #1
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    Telescopic sums

    Using the method of telescopic sums, how do I evaluate the following finite sums?

    S_1 = 3.2.3 + 3.3.4 + ... + 3(n+1)(n+2) + ... + 30906 using {n(n+1)(n+2)}

    S_2 = 2/(1.2.3) + 2/(2.3.4) + ... + 2/(n(n+1)(n+2)) + ... + 2/(1030200) using {-1/(n(n+1))}

    Thank you
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  2. #2
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    Quote Originally Posted by tsal15 View Post
    Using the method of telescopic sums, how do I evaluate the following finite sums?

    S_1 = 3.2.3 + 3.3.4 + ... + 3(n+1)(n+2) + ... + 30906 using {n(n+1)(n+2)}

    S_2 = 2/(1.2.3) + 2/(2.3.4) + ... + 2/(n(n+1)(n+2)) + ... + 2/(1030200) using {-1/(n(n+1))}

    Thank you
    S_2 has been asked and replied to in this thread: http://www.mathhelpforum.com/math-he...opic-sums.html. Any further correspondence regarding this sum should be asked there.
    Last edited by mr fantastic; October 25th 2008 at 07:35 PM.
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    Quote Originally Posted by mr fantastic View Post
    S_2 has been asked and replied to in this thread: http://www.mathhelpforum.com/math-he...opic-sums.html. Any further correspeondence regarding this sum should be asked there.
    Note that 30906 = 3(101)(102). Therefore

    S_1 = 3 \, [\, 2.3 + 3.4 + .... + (n+1)(n+2) + .... + (101)(102) \, ].


    Note that (n + 1)(n+2) = n^2 + 3n + 2.

    Therefore S_1 = 3 \sum_{n=1}^{100} (n+1)(n+2) = 3 \sum_{n=1}^{100} (n^2 + 3n + 2) = 3 \sum_{n=1}^{100} n^2 + 9 \sum_{n=1}^{100} n + 3 \sum_{n=1}^{100} 2 = 3 \sum_{n=1}^{100} n^2 + 9 \sum_{n=1}^{100} n + 3(2)(100) = \, ....

    You should have formulae you can use for getting \sum_{n=1}^{100} n and \sum_{n=1}^{100} n^2.
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