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anonymous

  • 5 years ago

use \[{{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} }\] to sum the series \[Sum_{n} = {2\over{1\times3}} - {4\over{3\times5}} + ... + {(-1)^{n-1} 2^n \over{(2n-1)(2n+1)}}\]

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  1. anonymous
    • 5 years ago
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    tried allsorts here but not got anywhere, difference method etc, there must be something i'm missing

  2. anonymous
    • 5 years ago
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    what is a good form

  3. anonymous
    • 5 years ago
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    that is exactly as the question was written

  4. anonymous
    • 5 years ago
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    Hmm I am nowhere near the solution but have you tried solving (n-1)th term and nth term together, I think you should try solving them maybe some good pattern emerges out, I am going to try it actually but the nth term is a little complicated and I might commit some mistakes. Note. Using this \({{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} }\)

  5. anonymous
    • 5 years ago
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    i wondered with there was a rearrangement that would give me a way to use difference method

  6. anonymous
    • 5 years ago
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    argh I am getting nowhere, If you want you can post this problem on "Meta-math" (Group on OpenStudy).

  7. anonymous
    • 5 years ago
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    BTW could the \(nth\) term be \(\large {(-1)^{n-1} 2n \over{(2n-1)(2n+1)}}\)?

  8. anonymous
    • 5 years ago
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    yes that is a typo

  9. anonymous
    • 5 years ago
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    \(2n\) instead of \(2^n\) can make the problem a lot simpler.

  10. anonymous
    • 5 years ago
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    Oh, so it is \(2n\). Thanks, maybe now I can get it...

  11. anonymous
    • 5 years ago
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    Okay, so things get much simpler now. I am gonna show \(nth\) and \((n-1)th\) only, we will see that every term cancels out except two terms in the whole expression i.e the last one and the first one. Let's assume the \(nth\) term to be negative, that makes \((n-1)th\) term positive. \[\frac{2(n-1)}{(2(n-1) -1)(2(n-1)+1)} - \frac{2n}{(2n-1)(2n+1)}\] \[\text{Using } {{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} }\] \[\frac{1}{2(2n-3)} + \frac{1}{2(2n-1)} - \frac{1}{2(2n-1)} + \frac{1}{2(2n+1)}\]

  12. anonymous
    • 5 years ago
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    I hope my logic is right.

  13. anonymous
    • 5 years ago
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    cool

  14. anonymous
    • 5 years ago
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    A similar thing must happen in the \((n-2)th\) term, what I mean is the \((n-2)th\) term must have a term of \(1\over 2(2n-3)\) which cancels out \(1\over 2(2n-3)\) that off \((n-1)th\) term. This whole process must continue until the first term (first term of the expanded form of first term of the series) i.e I think \(1\over 2\) Hmm final answer should look something like \(\frac{1}{2} + \frac{1}{2(2n-1)\) or \(\frac{1}{2} - \frac{1}{2(2n-1)\). My answer might be wrong but I think the concept is right.

  15. anonymous
    • 5 years ago
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    thanks for pointing in right direction

  16. anonymous
    • 5 years ago
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    uhhh, terrible I hate it when my LaTeX goes wrong. \[\frac{1}{2} \pm \frac{1}{2n+1}\] You're Welcome.

  17. anonymous
    • 5 years ago
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    It was a terrible typo which make the problem much harder, so now you have fixed the typo, I got the answer \( \huge \frac 12+ \frac{ (-1)^{(n-1)}}{2k+1} \)

  18. anonymous
    • 5 years ago
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    apologies, for that i'm not good with equation editing yet, i'm sure i'll get better, and its hard to edit once posted

  19. anonymous
    • 5 years ago
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    Oh it's okay, the lack of edit feature is really a bane while asking/answering this kind of questions. PS: *EDIT* The correct sum is \( \huge \frac 12+ \frac{ (-1)^{(k-1)}}{2k+1} \)

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