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anonymous

  • one year ago

Write the first three therms of the following series.Please help

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  1. Owlcoffee
    • one year ago
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    Do you have the series pattern given?

  2. anonymous
    • one year ago
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    Yes

  3. anonymous
    • one year ago
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    I will write it now

  4. anonymous
    • one year ago
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    \[\sum_{1}^{\infty} \frac{ 3n - 2 }{ n ^{2} + 1 }\]

  5. Owlcoffee
    • one year ago
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    A series is nothing more than a definable sum of terms, and it usually follows a pattern, and it is written inside the Sigma. \[\sum_{n=1}^{\infty} \frac{ 3n-2 }{ n^2+1 }\] we can find any term by limiting the sum to the desired term, in this case, it's 3: \[\sum_{n=1}^{n=3}\frac{ 3n-2 }{ n^2+1 }\] And, this just translates to: \[\sum_{n=1}^{n=3}\frac{ 3n-2 }{ n^2+1 }=\frac{ 3(1)-2 }{ (1)^2+1 }+\frac{ 3(2)-2 }{ (2)^2+1 } + \frac{ 3(3)-2 }{ (3)^2+1 }\]

  6. anonymous
    • one year ago
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    srry cant u just take a limit and divide the whole thing with the biggest power which in this case is n^2

  7. anonymous
    • one year ago
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    is it right what he jsut said?

  8. anonymous
    • one year ago
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    i am not sure if i am right i am pretty sure he is right

  9. Owlcoffee
    • one year ago
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    You take the limit when you desire to know if the series is convergent or divergent. When you want to find terms, you replace them on the actual pattern.

  10. anonymous
    • one year ago
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    I got another question

  11. anonymous
    • one year ago
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    Post it here or make open another topic?

  12. Owlcoffee
    • one year ago
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    Post it on another topic, so it doesn't get messy.

  13. anonymous
    • one year ago
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    ok

  14. anonymous
    • one year ago
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    oh ok srry i didnt read the question about three terms

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