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Math Help - partial sums of sequences

  1. #1
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    partial sums of sequences

    find the Sn of An = (1/n) -(1/n+2)
    i got the denominator n^2 + 3n + 2 but can't find the numerator
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  2. #2
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    \frac{1}{n}-\frac{1}{n+2}=\left( \frac{1}{n}-\frac{1}{n+1} \right)+\left( \frac{1}{n+1}-\frac{1}{n+2} \right), so \sum\limits_{n=1}^{m}{\left( \frac{1}{n}-\frac{1}{n+2} \right)}=\sum\limits_{n=1}^{m}{\left( \frac{1}{n}-\frac{1}{n+1} \right)}+\sum\limits_{n=1}^{m}{\left( \frac{1}{n+1}-\frac{1}{n+2} \right)}.

    can you finish?
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  3. #3
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    sorry, but no.
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  4. #4
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    okay then, try to look at what a telescoping sum is.
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  5. #5
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    Quote Originally Posted by confusedgirl View Post
    find the Sn of An = (1/n) -(1/n+2)
    i got the denominator n^2 + 3n + 2 but can't find the numerator
    Hi confusedgirl,

    You can write it this way...by placing in the values of n=1, 2, 3, 4, 5...

    A_n=\frac{1}{n}-\frac{1}{n+2}

    S_n=A_1+A_2+A_3+A_4+A_5+....A_{n-2}+A_{n-1}+A_n

    S_n=\left(\frac{1}{1}-\frac{1}{1+2}\right)+\left(\frac{1}{2}-\frac{1}{2+2}\right)+\left(\frac{1}{3}-\frac{1}{3+2}\right)+\left(\frac{1}{4}-\frac{1}{4+2}\right)......

    =\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+ .....+\left(\frac{1}{n-2}-\frac{1}{n}\right)+\left(\frac{1}{n-1}-\frac{1}{n+1}\right)+\left(\frac{1}{n}-\frac{1}{n+2}\right)

    If you examine this, you will see that virtually all of the terms cancel,
    due to the same terms having + and -.

    Which terms will be left standing?

    Also, what process did you use to calculate your original denominator?
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  6. #6
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    i'm sorry, but i think my question wasn't clear i need to get Sn where
    Sn = Sn-1 + An
    like
    S1 = 1-1/3 = 2/3
    S2 = 2/3 +1/2 -1/4 = 11/12
    S3= 21/20
    S4 = 34/30

    i really didn't use any process or technique to get the denominator but just figured it out by trial and error
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    okay then, try to look at what a telescoping sum is.
    ok
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  8. #8
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    1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+...\frac{1}{n-2}-\frac{1}{n}+\frac{1}{n-1}-\frac{1}{n+1}+\frac{1}{n}-\frac{1}{n+2}

    reduces to

    1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}=\frac{3}{2}-\left(\frac{1}{n+1}+\frac{1}{n+2}\right)

    since the last term to cancel is \frac{1}{n}

    as a term can only cancel a term 3 places to the right.

    =\frac{3}{2}-\frac{n+2+(n+1)}{(n+1)(n+2)}=\frac{3}{2}-\frac{2n+3}{(n^2+3n+2)}

    as all the other terms cancel.

    This will give you Sn for any n, without having to calculate a preceding sum.
    Last edited by Archie Meade; March 2nd 2010 at 04:39 PM. Reason: typo
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