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shaan_iitk
Group Title
Help needed... sum of series
= 1 + (3/4) + (7/16) + (15/64) + (31/256) + .. infinity
 2 years ago
 2 years ago
shaan_iitk Group Title
Help needed... sum of series = 1 + (3/4) + (7/16) + (15/64) + (31/256) + .. infinity
 2 years ago
 2 years ago

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tanusingh Group TitleBest ResponseYou've already chosen the best response.0
firstly is it a g.p?
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
okk I got this much that = 1 + (2^2 1)/2^2 + (2^3 1)/(2^4) + (2^4  1)/2^16 + ... = \[2^2 * \sum_{0}^{\infty} (2^n1)/(2^2)^n\] Now what ????
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
its not a gp.. its a combination .. think someone with little practice can answer this .. I have lost touch and hence have forgot the method to sum these..
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
also .. my formula is wrong .. I think it would be somewhat like (2^(n+1)  1)/(2^(2(2n2))
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{2^{n} 1}{2^{2n}} = \frac{1}{2^{n}}\frac{1}{2^{2n}} =(\frac{1}{2})^{n}  (\frac{1}{4})^{n}\] \[\rightarrow 4*[\sum_{1}^{\infty}(\frac{1}{2})^{n}  \sum_{1}^{\infty}(\frac{1}{4})^{n}]\]
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
no i think your formula worked if you start at n=1
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
@dumbcow .. no I mistyped the formula .. look at the series ... you won't get 7/16 .. or 31/256 ... it is a little more complicated...
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
(2^5 1)/(2^2)^4
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
got it ... thanks .. yeah my formula is just slightly of.. but thanks anyways.. :)
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.1
\(\large \frac 23 \) isn't ?
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
it cannot be 2/3 because it is (1 + 3/4 + ... ) and all terms are positive
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
no thats right for the summation part. multiply by 4 and you get , sum = 8/3
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
yeah ... it wd be 8/3 ..
 2 years ago

shaan_iitk Group TitleBest ResponseYou've already chosen the best response.0
thanks both of you.. :)
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.1
yes it should be 8/3, I summed up dumcow's generalization, but the actualy generalization should be \( \huge \sum\limits_{k=0}^\infty \frac {2^{k+1} 1}{2^{2k}}=\frac 83\)
 2 years ago

FoolForMath Group TitleBest ResponseYou've already chosen the best response.1
It wasn't dumcow's I didn't read the thread (fully). My apologies.
 2 years ago
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