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Need to use the binomal therom to expand (1+2x)/(1-2x) Not sure where to start to get it into the right format.

Mathematics
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\[(1+2x)/(1-2x)\]
Oh. Need to go up too and including the term \[x ^{2}\]
so its x^2+2x+1/1-2x?

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Other answers:

Could you explain the steps? thank you very much for your timee!
Is that \[(1+2x) \times (1-2x)\] or \[\frac{(1+2x)}{(1-2x)}\]?
I have been trying to do those Fractions but am unable to do it! Soo fustrating! Yes it is the the (1+2x) over (1-2x)
Okay give me a minute to work this out :)
If it is any help the answer is \[(1+2x)(1+2x+4x ^{2})\]
(1+2x) over (1-2x) can never be equal to (1+2x)(1+2x+4x²) and you can assure that by substitution.
\[(1+2x)(1-2x)^{-1}=(1+2x)(1^{-1}+-1*1^{-1-1}*-2x+\frac{-1(-1-1)}{2} (1)^{-1-2}(2x)^2+...)\]\[=(1+2x)(1^{-1}+2x+\frac{-1(-2)}{2} (1)^{-3}*4x^2+...)\]\[=(1+2x)(1+2x+\frac{2}{2} *1*4x^2+...)\]\[=(1+2x)(1+2x+4x^2+...)\] This is from this rule \[(a+b)^n=a^n+na^{n-1}b+\frac{n(n-1)}{2}a^{n-2}b^2+....\] Sorry it took so long :D
either the answer is wrong or you forgot something when posting the problem i agree with y2o2
http://mathworld.wolfram.com/NegativeBinomialSeries.html
@zed yes it becomes an infinite sum..is that the solution they are looking for? their answer stops after 4x^2
Yes they only had to do the terms until it reaches x^2 power
ok thanks for clearing it up :)
Thank you very much guys! Clears things up!

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