anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[(1+2x)/(1-2x)\]
anonymous
  • anonymous
Oh. Need to go up too and including the term \[x ^{2}\]
King
  • King
so its x^2+2x+1/1-2x?

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

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