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quixie quixie
wrote...
8 years ago
Compute the sum and the limit of the sum as n -> infinity.

\({\sum_{n=1}^n \frac{1}{n}[13(i/n)^2-9(i/n)]}\)

a. Sum = \(\frac{13(n+1)(2n+1)}{6n^2}-\frac{9(n+1)}{2n}\) ; limit of sum as n -> infinity is -1/6

b. Sum = \(\frac{(n+1)(2n+1)}{13n^2}-\frac{n+1}{9n}\) ; limit of sum as n -> infinity is -17/117

c. Sum = \(\frac{13(n+1)(2n+1)}{6n^2}-\frac{9(n+1)}{2n}\) ; limit of sum as n -> infinity is 0

d. Sum = \(\frac{(n+1)(2n+1)}{13n^2}-\frac{n+1}{9n}\) ; limit of sum as n -> infinity is 17/117
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