Question

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Answer 1

Think of the first 1 as 1/1
What are the denominators?

Answer 2

55?

Answer 3

Term 1 has denominator 1.
Term 2 has denominator 4.
Term 3 has denominator 9.
Term 4 has denominator 16.
etc.
Do you see a connection between the term number and the actual denominator?

Answer 4

oh yeah

Answer 5

\[1+\frac{ 3 }{ 4 }+\frac{ 5 }{ 9 }+\frac{ 7 }{ 16 }+\frac{ 9 }{ 25 }+...\]

Answer 6

For clarity. LOL

Answer 7

yeah thts correct

Answer 8

Okay, the expression in sigma notation is \[\sum_{n=1}^{\infty} \frac{ 2n-1 }{ n^{2} }\]

Answer 9

Sorry, I wrote the answer immediately. To explain, the denominators are just the squares of the integers. The numerators are odd integers, and odd integers are of the form 2n-1, where n is an integer.

Answer 10

ohhh lol that makes sense thank you!

Answer 11

In your case, if the sum is only the first 5 terms you showed, then it should read:

\(\Large \sum_{n=1}^{5} \frac{ 2n-1 }{ n^{2} }\)