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

First find ( S_m ): [ S_m = 3\cdot\frac{m(m+1)(2m+1)}{6} + 5\cdot\frac{m(m+1)}{2} + 2m = \frac{m(m+1)(2m+1)}{2} + \frac{5m(m+1)}{2} + 2m = \frac{m}{2}\bigl[(m+1)(2m+1) + 5(m+1) + 4\bigr] = \frac{m}{2}(2m^2 + m + 2m + 1 + 5m + 5 + 4) = \frac{m}{2}(2m^2 + 8m + 10) = m(m^2 + 4m + 5) = m(m+1)(m+4). ]

Now put ( m = 2n ): [ S_{2n} = 2n(2n+1)(2n+4) = 2n(2n+1)\cdot 2(n+2) = 4n(2n+1)(n+2). ]

Answer: ( 4n(2n+1)(n+2) )

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

[ S_m = \frac{m(m+1)(2m+1)}{6} + \frac{m(m+1)}{2} - 2m = \frac{m(m+1)(2m+1)}{6} + \frac{3m(m+1)}{6} - \frac{12m}{6} = \frac{m}{6}\bigl[(m+1)(2m+1) + 3(m+1) - 12\bigr] = \frac{m}{6}(2m^2 + m + 2m + 1 + 3m + 3 - 12) = \frac{m}{6}(2m^2 + 6m - 8) = \frac{m}{3}(m^2 + 3m - 4) = \frac{m(m+4)(m-1)}{3}. ]

Put ( m = 2n ): [ S_{2n} = \frac{2n(2n+4)(2n-1)}{3} = \frac{2n\cdot 2(n+2)(2n-1)}{3} = \frac{4n(n+2)(2n-1)}{3}. ]

Answer: ( \dfrac{4n(n+2)(2n-1)}{3} )