pythagoras123
  • pythagoras123
See attachment:
Mathematics
jamiebookeater
  • jamiebookeater
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pythagoras123
  • pythagoras123
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anonymous
  • anonymous
oh you lazy
AccessDenied
  • AccessDenied
You could probably just sequentially find common denominators between the first terms of the sequence, compute the sum or difference of those numbers, and then move on to the next number carrying the number you found, repeat. e.g. 1 - 5/6 6/6 - 5/6 1/6 (1/6) + 7/12 2/12 + 7/12 9/12 and so on

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anonymous
  • anonymous
3/5
pythagoras123
  • pythagoras123
Dear santistebanc, I already know the answer, I just don't know how to do it in a faster way, that's all.
anonymous
  • anonymous
|dw:1333869325834:dw| maybe i'll just use my calculator... sorry...
anonymous
  • anonymous
Set up the summation: \[1+\sum_{n=2}^{9} \frac {(1+2n)(-1)^{n+1}} {(n+1)n} \]
anonymous
  • anonymous
From there you can use the Alternating Series Estimation Theorem.

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