Mathematics
OpenStudy (anonymous):

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

OpenStudy (anonymous):

tried allsorts here but not got anywhere, difference method etc, there must be something i'm missing

OpenStudy (anonymous):

what is a good form

OpenStudy (anonymous):

that is exactly as the question was written

OpenStudy (anonymous):

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 $${{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} }$$

OpenStudy (anonymous):

i wondered with there was a rearrangement that would give me a way to use difference method

OpenStudy (anonymous):

argh I am getting nowhere, If you want you can post this problem on "Meta-math" (Group on OpenStudy).

OpenStudy (anonymous):

BTW could the $$nth$$ term be $$\large {(-1)^{n-1} 2n \over{(2n-1)(2n+1)}}$$?

OpenStudy (anonymous):

yes that is a typo

OpenStudy (anonymous):

$$2n$$ instead of $$2^n$$ can make the problem a lot simpler.

OpenStudy (anonymous):

Oh, so it is $$2n$$. Thanks, maybe now I can get it...

OpenStudy (anonymous):

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. $\frac{2(n-1)}{(2(n-1) -1)(2(n-1)+1)} - \frac{2n}{(2n-1)(2n+1)}$ $\text{Using } {{1 \over 2}\over{r-1}} + {{1 \over 2}\over{r+1}} = {r \over{(r-1) (r+1)} }$ $\frac{1}{2(2n-3)} + \frac{1}{2(2n-1)} - \frac{1}{2(2n-1)} + \frac{1}{2(2n+1)}$

OpenStudy (anonymous):

I hope my logic is right.

OpenStudy (anonymous):

cool

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

thanks for pointing in right direction

OpenStudy (anonymous):

uhhh, terrible I hate it when my LaTeX goes wrong. $\frac{1}{2} \pm \frac{1}{2n+1}$ You're Welcome.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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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