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anonymous

  • one year ago

Sigma (summation) notation question. Will post in comment

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  1. anonymous
    • one year ago
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    \[\sum_{k = 0}^{4}\frac{ 1 }{ k^2 + 1 }\] I rewrote it as: \[\sum_{k = 0}^{4} (k^2 + 1)^{-1}\] do I use now the formula \[\frac{ n(n+1)(2n+1) }{ 6 }\] for the expression (k^2 + 1)^-1 ?

  2. anonymous
    • one year ago
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    I think that what I wrote in the previous step is incorrect; I don't think the formula can be applied. It can be solved with brute force. \[\frac{ 1 }{ 0^2 + 1} + \frac{ 1 }{ 1^2 + 1} + \frac{ 1 }{ 2^2 + 1} + \frac{ 1 }{ 3^2 + 1} + \frac{ 1 }{ 4^2 + 1}\] \[= \frac{ 1 }{ 0 + 1 } + \frac{ 1 }{ 1 + 1 } + \frac{ 1 }{ 4 + 1 } + \frac{ 1 }{ 9 + 1 } + \frac{ 1 }{ 16 + 1 }\] \[= \frac{ 1 }{ 1 } + \frac{ 1 }{ 2 } + \frac{ 1 }{ 5 } + \frac{ 1 }{ 10 } + \frac{ 1 }{ 17 } =\frac{ 158 }{ 85 } \]

  3. anonymous
    • one year ago
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    you are correct in the second solution :)

  4. anonymous
    • one year ago
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    @Halmos I'm I correct that the summation formula won't work? The formulas can be found @ https://mathwiz.uwstout.edu/video/math-156/ch5s1.html 4th video down (It's not necessary to watch the video, it's on the web page).

  5. anonymous
    • one year ago
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    the formula u had of k^2 won't work but there is a formula called zeta function and like wise things, this is advanced mathematics don't think about it :P

  6. anonymous
    • one year ago
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    Thanks.

  7. anonymous
    • one year ago
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    np :)

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