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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Do you have the series pattern given?
Yes
I will write it now

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Other answers:

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

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