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Need to use the binomal therom to expand (1+2x)/(12x) Not sure where to start to get it into the right format.
 2 years ago
 2 years ago
Need to use the binomal therom to expand (1+2x)/(12x) Not sure where to start to get it into the right format.
 2 years ago
 2 years ago

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jpknegtelBest ResponseYou've already chosen the best response.0
Oh. Need to go up too and including the term \[x ^{2}\]
 2 years ago

jpknegtelBest ResponseYou've already chosen the best response.0
Could you explain the steps? thank you very much for your timee!
 2 years ago

ZedBest ResponseYou've already chosen the best response.3
Is that \[(1+2x) \times (12x)\] or \[\frac{(1+2x)}{(12x)}\]?
 2 years ago

jpknegtelBest ResponseYou've already chosen the best response.0
I have been trying to do those Fractions but am unable to do it! Soo fustrating! Yes it is the the (1+2x) over (12x)
 2 years ago

ZedBest ResponseYou've already chosen the best response.3
Okay give me a minute to work this out :)
 2 years ago

jpknegtelBest ResponseYou've already chosen the best response.0
If it is any help the answer is \[(1+2x)(1+2x+4x ^{2})\]
 2 years ago

y2o2Best ResponseYou've already chosen the best response.1
(1+2x) over (12x) can never be equal to (1+2x)(1+2x+4x²) and you can assure that by substitution.
 2 years ago

ZedBest ResponseYou've already chosen the best response.3
\[(1+2x)(12x)^{1}=(1+2x)(1^{1}+1*1^{11}*2x+\frac{1(11)}{2} (1)^{12}(2x)^2+...)\]\[=(1+2x)(1^{1}+2x+\frac{1(2)}{2} (1)^{3}*4x^2+...)\]\[=(1+2x)(1+2x+\frac{2}{2} *1*4x^2+...)\]\[=(1+2x)(1+2x+4x^2+...)\] This is from this rule \[(a+b)^n=a^n+na^{n1}b+\frac{n(n1)}{2}a^{n2}b^2+....\] Sorry it took so long :D
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.0
either the answer is wrong or you forgot something when posting the problem i agree with y2o2
 2 years ago

ZedBest ResponseYou've already chosen the best response.3
http://mathworld.wolfram.com/NegativeBinomialSeries.html
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.0
@zed yes it becomes an infinite sum..is that the solution they are looking for? their answer stops after 4x^2
 2 years ago

ZedBest ResponseYou've already chosen the best response.3
Yes they only had to do the terms until it reaches x^2 power
 2 years ago

dumbcowBest ResponseYou've already chosen the best response.0
ok thanks for clearing it up :)
 2 years ago

jpknegtelBest ResponseYou've already chosen the best response.0
Thank you very much guys! Clears things up!
 2 years ago
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