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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

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  1. tanusingh Group Title
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    firstly is it a g.p?

    • 2 years ago
  2. shaan_iitk Group Title
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    okk I got this much that = 1 + (2^2 -1)/2^2 + (2^3 -1)/(2^4) + (2^4 - 1)/2^16 + ... = \[2^2 * \sum_{0}^{\infty} (2^n-1)/(2^2)^n\] Now what ????

    • 2 years ago
  3. shaan_iitk Group Title
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    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
  4. shaan_iitk Group Title
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    also .. my formula is wrong .. I think it would be somewhat like (2^(n+1) - 1)/(2^(2(2n-2))

    • 2 years ago
  5. dumbcow Group Title
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    \[\frac{2^{n} -1}{2^{2n}} = \frac{1}{2^{n}}-\frac{1}{2^{2n}} =(\frac{1}{2})^{n} - (\frac{1}{4})^{n}\] \[\rightarrow 4*[\sum_{1}^{\infty}(\frac{1}{2})^{n} - \sum_{1}^{\infty}(\frac{1}{4})^{n}]\]

    • 2 years ago
  6. dumbcow Group Title
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    no i think your formula worked if you start at n=1

    • 2 years ago
  7. shaan_iitk Group Title
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    @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
  8. shaan_iitk Group Title
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    (2^5 -1)/(2^2)^4

    • 2 years ago
  9. shaan_iitk Group Title
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    got it ... thanks .. yeah my formula is just slightly of.. but thanks anyways.. :)

    • 2 years ago
  10. FoolForMath Group Title
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    \(\large \frac 23 \) isn't ?

    • 2 years ago
  11. shaan_iitk Group Title
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    it cannot be 2/3 because it is (1 + 3/4 + ... ) and all terms are positive

    • 2 years ago
  12. dumbcow Group Title
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    no thats right for the summation part. multiply by 4 and you get , sum = 8/3

    • 2 years ago
  13. shaan_iitk Group Title
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    yeah ... it wd be 8/3 ..

    • 2 years ago
  14. shaan_iitk Group Title
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    thanks both of you.. :)

    • 2 years ago
  15. FoolForMath Group Title
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    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
  16. FoolForMath Group Title
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    It wasn't dumcow's I didn't read the thread (fully). My apologies.

    • 2 years ago
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