(i) ( u_n = 5n^2 + 2n + 3 )

[ S_n = 5\sum k^2 + 2\sum k + 3\sum 1 = 5\cdot\frac{n(n+1)(2n+1)}{6} + 2\cdot\frac{n(n+1)}{2} + 3n = \frac{5n(n+1)(2n+1)}{6} + n(n+1) + 3n = \frac{n}{6}\Bigl[5(n+1)(2n+1) + 6(n+1) + 18\Bigr] = \frac{n}{6}(10n^2 + 5n + 10n + 5 + 6n + 6 + 18) = \frac{n}{6}(10n^2 + 21n + 29) = \frac{n(10n^2 + 21n + 29)}{6}. ]

Answer: ( \dfrac{n(10n^2 + 21n + 29)}{6} )

(ii) ( u_n = n^2 + 2n - 3 )

[ S_n = \sum k^2 + 2\sum k - 3\sum 1 = \frac{n(n+1)(2n+1)}{6} + 2\cdot\frac{n(n+1)}{2} - 3n = \frac{n(n+1)(2n+1)}{6} + n(n+1) - 3n = \frac{n}{6}\Bigl[(n+1)(2n+1) + 6(n+1) - 18\Bigr] = \frac{n}{6}(2n^2 + n + 2n + 1 + 6n + 6 - 18) = \frac{n}{6}(2n^2 + 9n - 11) = \frac{n(2n^2 + 9n - 11)}{6}. ]

Answer: ( \dfrac{n(2n^2 + 9n - 11)}{6} )