A community for students.
Here's the question you clicked on:
 0 viewing
shaan_iitk
 4 years ago
Help needed... sum of series
= 1 + (3/4) + (7/16) + (15/64) + (31/256) + .. infinity
shaan_iitk
 4 years ago
Help needed... sum of series = 1 + (3/4) + (7/16) + (15/64) + (31/256) + .. infinity

This Question is Closed

shaan_iitk
 4 years ago
Best ResponseYou've already chosen the best response.0okk 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 ????

shaan_iitk
 4 years ago
Best ResponseYou've already chosen the best response.0its 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..

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

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

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

shaan_iitk
 4 years ago
Best 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...

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

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

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

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

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

shaan_iitk
 4 years ago
Best ResponseYou've already chosen the best response.0thanks both of you.. :)

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

FoolForMath
 4 years ago
Best ResponseYou've already chosen the best response.1It wasn't dumcow's I didn't read the thread (fully). My apologies.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.