(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) )