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Math Help - Sum of Series

  1. #1
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    Sum of Series

    Find the sum of the first n terms of the series:

    (i) 1.3.5 + 3.5.7 + 5.7.9 + ... + (2n - 1)(2n + 1)(2n +3)

    (ii) 1/(1.3.5) + 1/(3.5.7) + 1/(5.7.9) + ... + 1/[(2n - 1)(2n + 1)(2n +3)]

    I have tried many methods, but with no avail... how can i solve this while displaying working and explanations??
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  2. #2
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    Quote Originally Posted by cedricc View Post
    Find the sum of the first n terms of the series:

    (i) 1.3.5 + 3.5.7 + 5.7.9 + ... + (2n - 1)(2n + 1)(2n +3)

    (ii) 1/(1.3.5) + 1/(3.5.7) + 1/(5.7.9) + ... + 1/[(2n - 1)(2n + 1)(2n +3)]

    I have tried many methods, but with no avail... how can i solve this while displaying working and explanations??
    i duno how to do the sum sign, so im sorry. so when i do the sum sign that means r=1 to n.

    but i) (2n - 1)(2n + 1)(2n +3) times that out.
    which comes to 8n^3+12n^2-2n-3.

    so for the sum from r=1 to n = 8\sum r^3 + 12\sum r^2 -2 \sum r -3r.

    then using the fact

    \sum r= \frac{1}{2}n(n+1)
    \sum r^2= \frac{1}{6}n(n+1)(2n+1)
    \sum r^3= \frac{1}{4}n^2(n+1)^2

    sub this in 8\sum r^3 + 12\sum r^2 -2 \sum r -3r.

    do you understand?
    Last edited by BabyMilo; April 25th 2010 at 03:17 AM.
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  3. #3
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    i don't really follow, sorry...
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  4. #4
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    Quote Originally Posted by cedricc View Post
    i don't really follow, sorry...
    which part?
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  5. #5
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    Actually, nevermind, i got it ^^ Thanks!
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  6. #6
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    Go here: The On-Line Encyclopedia of Integer Sequences
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  7. #7
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    Quote Originally Posted by cedricc View Post
    (ii) 1/(1.3.5) + 1/(3.5.7) + 1/(5.7.9) + ... + 1/[(2n - 1)(2n + 1)(2n +3)]
    this part is much interesting, it can be solved in few lines:

    \frac{1}{(2n-1)(2n+1)(2n+3)}=\frac{1}{4}\cdot \frac{(2n+3)-(2n-1)}{(2n-1)(2n+1)(2n+3)}, thus the general term equals \frac{1}{4}\left( \frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)} \right). Now let a_n=\frac{1}{(2n-1)(2n+1)} and see that a_{n+1}=\frac{1}{(2n+1)(2n+3)}, so \frac{1}{4}\sum\limits_{n=1}^{m}{\left( \frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n+1)(2n+3)} \right)} is a telescoping sum.
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