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Math Help - Series...

  1. #1
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    Series...

    Find the sum of:

    1/(1*2*3) + 1(2*3*4) + 1/(3*4*5) + ... + 1/(n(n+1)(n+2))
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  2. #2
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    Quote Originally Posted by bearej50 View Post
    Find the sum of:


    1/(1*2*3) + 1(2*3*4) + 1/(3*4*5) + ... + 1/(n(n+1)(n+2))


    Telescopic series: \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\Longrightarrow \frac{1}{n(n+1)(n+2)}=\left(\frac{1}{n}-\frac{1}{n+1}\right)\frac{1}{n+2}= \frac{1}{n(n+2)} -\frac{1}{(n+1)(n+2)} =\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)-\left(\frac{1}{n+1}-\frac{1}{n+2}\right) .

    I think you can take it from here

    Tonio
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  3. #3
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    note that in tonio's solution the first term telescopes not so fast than the second term which does faster.

    another way on doing this is to see that \frac{1}{n(n+1)(n+2)}=\frac{1}{2}\cdot \frac{n+2-n}{n(n+1)(n+2)}=\frac{1}{2}\left( \frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)} \right), and this one telescopes easily.
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    note that in tonio's solution the first term telescopes not so fast than the second term which does faster.

    another way on doing this is to see that \frac{1}{n(n+1)(n+2)}=\frac{1}{2}\cdot \frac{n+2-n}{n(n+1)(n+2)}=\frac{1}{2}\left( \frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)} \right), and this one telescopes easily.

    Much better than my solution.

    Tonio
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