Let f (x) = x ^ 2 + X, when x belongs to [n, N + 1], (n belongs to positive integer), the number of all integer values of F (x) is g (n) Let an = (2n ^ 3 + 3N ^ 2) / g (n), n be a positive integer, and then find s (n)

F (n) = n ^ 2 + NF (n + 1) = n ^ 2 + 2n + 1 + N + 1 = the number of integers between n ^ 2 + 3N + 2, G (n) = f (n + 1) - f (n) + 1 = 2n + 3an = (2n ^ 3 + 3N ^ 2) / (2n + 3) = n ^ 2 (an + 3) / (2n + 3) = n ^ 2, so Sn = 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + n^2 = n(n+1)(2n+1)/6...

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