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anonymous
 5 years ago
use \[{{1 \over 2}\over{r1}} + {{1 \over 2}\over{r+1}} = {r \over{(r1) (r+1)} }\] to sum the series
\[Sum_{n} = {2\over{1\times3}}  {4\over{3\times5}} + ... + {(1)^{n1} 2^n \over{(2n1)(2n+1)}}\]
anonymous
 5 years ago
use \[{{1 \over 2}\over{r1}} + {{1 \over 2}\over{r+1}} = {r \over{(r1) (r+1)} }\] to sum the series \[Sum_{n} = {2\over{1\times3}}  {4\over{3\times5}} + ... + {(1)^{n1} 2^n \over{(2n1)(2n+1)}}\]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tried allsorts here but not got anywhere, difference method etc, there must be something i'm missing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is exactly as the question was written

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm I am nowhere near the solution but have you tried solving (n1)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 \({{1 \over 2}\over{r1}} + {{1 \over 2}\over{r+1}} = {r \over{(r1) (r+1)} }\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i wondered with there was a rearrangement that would give me a way to use difference method

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0argh I am getting nowhere, If you want you can post this problem on "Metamath" (Group on OpenStudy).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0BTW could the \(nth\) term be \(\large {(1)^{n1} 2n \over{(2n1)(2n+1)}}\)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\(2n\) instead of \(2^n\) can make the problem a lot simpler.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, so it is \(2n\). Thanks, maybe now I can get it...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, so things get much simpler now. I am gonna show \(nth\) and \((n1)th\) 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 \((n1)th\) term positive. \[\frac{2(n1)}{(2(n1) 1)(2(n1)+1)}  \frac{2n}{(2n1)(2n+1)}\] \[\text{Using } {{1 \over 2}\over{r1}} + {{1 \over 2}\over{r+1}} = {r \over{(r1) (r+1)} }\] \[\frac{1}{2(2n3)} + \frac{1}{2(2n1)}  \frac{1}{2(2n1)} + \frac{1}{2(2n+1)}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I hope my logic is right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A similar thing must happen in the \((n2)th\) term, what I mean is the \((n2)th\) term must have a term of \(1\over 2(2n3)\) which cancels out \(1\over 2(2n3)\) that off \((n1)th\) 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 \(\frac{1}{2} + \frac{1}{2(2n1)\) or \(\frac{1}{2}  \frac{1}{2(2n1)\). My answer might be wrong but I think the concept is right.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thanks for pointing in right direction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0uhhh, terrible I hate it when my LaTeX goes wrong. \[\frac{1}{2} \pm \frac{1}{2n+1}\] You're Welcome.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It was a terrible typo which make the problem much harder, so now you have fixed the typo, I got the answer \( \huge \frac 12+ \frac{ (1)^{(n1)}}{2k+1} \)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0apologies, for that i'm not good with equation editing yet, i'm sure i'll get better, and its hard to edit once posted

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh 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 \( \huge \frac 12+ \frac{ (1)^{(k1)}}{2k+1} \)
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