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9 August, 13:30

Calculus 2: Sum from n=1 to infinity of (n^4 + ((-1) ^n) * n^2) / ((-n) ^5 + 3 (n^3))

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  1. 9 August, 13:54
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    (n^4 + ((-1) ^n) * n^2) / ((-n) ^5 + 3 (n^3))

    n^4 + - 1 / - 1^5 + 3n^3

    n^4 + - 1^n / - 1^5 + 3^3

    n^4 / - 1 + 3n^3

    n^4 / - 1 + 3x1^3

    n/-1 + 3n

    1/-1+3

    n = 2
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