anonymous
  • anonymous
Write the first three therms of the following series.Please help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Owlcoffee
  • Owlcoffee
Do you have the series pattern given?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
I will write it now

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anonymous
  • anonymous
\[\sum_{1}^{\infty} \frac{ 3n - 2 }{ n ^{2} + 1 }\]
Owlcoffee
  • Owlcoffee
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: \[\sum_{n=1}^{n=3}\frac{ 3n-2 }{ n^2+1 }=\frac{ 3(1)-2 }{ (1)^2+1 }+\frac{ 3(2)-2 }{ (2)^2+1 } + \frac{ 3(3)-2 }{ (3)^2+1 }\]
anonymous
  • anonymous
srry cant u just take a limit and divide the whole thing with the biggest power which in this case is n^2
anonymous
  • anonymous
is it right what he jsut said?
anonymous
  • anonymous
i am not sure if i am right i am pretty sure he is right
Owlcoffee
  • Owlcoffee
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.
anonymous
  • anonymous
I got another question
anonymous
  • anonymous
Post it here or make open another topic?
Owlcoffee
  • Owlcoffee
Post it on another topic, so it doesn't get messy.
anonymous
  • anonymous
ok
anonymous
  • anonymous
oh ok srry i didnt read the question about three terms

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