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Yahoo!

  • 3 years ago

\[Sn = \sum_{k=1}^{n}\frac{ k^2 }{ 1+n^3 }\] , then \[\lim_{n \rightarrow \infty }Sn = ?\]

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  1. joemath314159
    • 3 years ago
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    You can factor the 1+n^3 out of the denominator, then use the fact that:\[\sum_{k=0}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\]

  2. Yahoo!
    • 3 years ago
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    Yup..i got it...thxxx

  3. Yahoo!
    • 3 years ago
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    is it 1/3 ? @joemath314159

  4. joemath314159
    • 3 years ago
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    yes thats correct :)

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