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Sequences/Series question

  1. Nov 1, 2005 #1
    I don't know how the latex coding works sorry...

    I have to use the partial fraction technique on 1/(4k^2 - 1)...
    So far so good and I get 1 / 2(2k-1) - 1 / 2(2k+1), is this correct?


    I now need to show that if E is the sum of terms where k=1 and n is the unknow.

    E 1 / 4k^2 - 1 = n / 2n + 1

    Please help :cry:
     
    Last edited: Nov 1, 2005
  2. jcsd
  3. Nov 1, 2005 #2

    verty

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    Homework Helper

    For the first part, notice: 4k^2 - 1 = (2k-1)(2k+1)

    I don't understand the second part, it makes no sense. If it was asking for the value of 'n' it would make sense. There's no reason to ask whether E can be expressed in that form when k = 1, I can't see why a teacher would want to ask it.

    Also, what does this have to do with sequences/series?

    Edit: I see about the sigma notation now, totally misunderstood the question the first time, don't know how to solve it though.
     
    Last edited: Nov 1, 2005
  4. Nov 1, 2005 #3

    HallsofIvy

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    [tex]\Sigma_{k=1}^n\frac{1}{4k^2-1}= \frac{n}{2n+1}[/tex]
    Is that what you mean? Even if you can't use latex, please use parentheses! Most people would read 1/4k^2-1 as (1/4)k^2- 1 or 1/(4k^2)- 1 rather than 1/(4k^2-1).
    Yes, your partial fraction decomposition is correct. To see how it applies to the sum, write out a few of the terms using that decomposition:
    doing, say k= 1 to 4:
    k= 1 [tex]\frac{1}{4-1}= \frac{1}{2}- \frac{1}{6}[/tex]
    k= 2 [tex]\frac{1}{16-1}= \frac{1}{6}- \frac{1}{10}[/tex]
    k= 3 [tex]\frac{1}{36-1}= \frac{1}{10}- \frac{1}{14}[/tex]
    k= 4 [tex]\frac{1}{63}= \frac{1}{14}- \frac{1}{18}[/tex]
    Do you see what happens when you add those?
     
  5. Nov 1, 2005 #4

    Yes thanks I have now completed the exercise :-). Thanks!
     
  6. Nov 1, 2005 #5
    Exercise done thanks!
     
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