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Using partial fractions, we can observe that:
(2n−1)(2n+1)232n=2n−14−2n+14+(2n+1)28
Now ∑n=1∞(2n−14−2n+14) is a telescoping sum, whose value is 4. So there is just one more sum to calculate:
n=1∑∞(2n+1)28=8(k=1∑∞k21−n=1∑∞(2n)21−121)=68π2−4∗68π2−8=π2−8
Two different Countries, two different Timezones but then also you two, writing the same solution of the same question at the same time. Wow! These is possible on Brilliant only. Cheers....
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Using partial fractions, we can observe that: (2n−1)(2n+1)232n=2n−14−2n+14+(2n+1)28
Now ∑n=1∞(2n−14−2n+14) is a telescoping sum, whose value is 4. So there is just one more sum to calculate: n=1∑∞(2n+1)28=8(k=1∑∞k21−n=1∑∞(2n)21−121)=68π2−4∗68π2−8=π2−8
Therefore, 4+32n=1∑∞(2n−1)(2n+1)2n=4+(4+π2−8)=π2QED
I'm curious how you got this from the series of 1−x2x2?
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so ,i used the expansion of 1−x2x2 which is n=1∑∞4n−1(2n−2n−2)x2n integrated both sides then put x=sint and i got
n=1∑∞4n−1(2n−2n−2)2n+1sin2n+1t=2t−sintcost
then i integrated from 0 to 2π and got with a lot of rearranging and simplifying and comparing with other sums i get π2=4+32n=1∑∞(2n−1)(2n+1)2n
this is how it looked before all the comparing and substituting
16π2−41=21n=1∑∞4n−1(2n+1)Γ(n+23)Γ(n+1)π(2n−2n−1)
this was a practice problem in my textbook,so i had faith in solving it ;),it was a bit tedious though
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Wow, looks like a lot of work! Nice job though!
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You beated me while writing the solution.
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Two different Countries, two different Timezones but then also you two, writing the same solution of the same question at the same time. Wow! These is possible on Brilliant only. Cheers....
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Interesting, we both had the same solution at the same time
i'll post how once i get back from school :)
4+32n=1∑∞(2n−1)(2n+1)2n4+32n=1∑∞(8(2n−1)1−8(2n+1)1+4(2n+1)21)4+n=1∑∞((2n−1)4−(2n+1)4+(2n+1)28)4+⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛4+34+54+…−34−54−…+8n=1∑∞n21−n=1∑∞(2n)21−1=6π2−24π2−1=8π2−1n=1∑∞(2n+1)21⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞8+8(8π2−1)8+π2−8=π2
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Very neat solution! +1
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Thank you very much ! :-)