(i) ( 1\times 3 + 2\times 5 + 3\times 7 + \cdots )
The ( n^{th} ) term is [ u_n = n(2n+1) = 2n^2 + n. ]
Sum to ( n ) terms: [ S_n = \sum_{k=1}^{n} (2k^2 + k) = 2\cdot\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} = \frac{n(n+1)}{6}\bigl[2(2n+1) + 3\bigr] = \frac{n(n+1)(4n+5)}{6}. ]
Answer: ( \dfrac{n(n+1)(4n+5)}{6} )