(i) ( 1^2 - 2^2 + 3^2 - 4^2 + \cdots + (2n-1)^2 - (2n)^2 )

Group terms in pairs: [ \bigl[(2k-1)^2 - (2k)^2\bigr] = (4k^2 - 4k + 1) - 4k^2 = -4k + 1. ]

There are ( n ) such pairs, so [ S_n = \sum_{k=1}^{n} (1 - 4k) = \sum_{k=1}^{n} 1 - 4\sum_{k=1}^{n} k = n - 4\cdot\frac{n(n+1)}{2} = n - 2n(n+1) = n\bigl[1 - 2(n+1)\bigr] = n(1 - 2n - 2) = n(-2n - 1) = -n(2n+1). ]

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

(ii) ( \dfrac{1^2}{1} + \dfrac{1^2+2^2}{2} + \dfrac{1^2+2^2+3^2}{3} + \cdots ) to ( n ) terms

The ( r^{th} ) term is [ u_r = \frac{1}{r}\sum_{k=1}^{r} k^2 = \frac{1}{r}\cdot\frac{r(r+1)(2r+1)}{6} = \frac{(r+1)(2r+1)}{6} = \frac{2r^2 + 3r + 1}{6}. ]

[ S_n = \frac{1}{6}\sum_{r=1}^{n} (2r^2 + 3r + 1) = \frac{1}{6}\Bigl[2\cdot\frac{n(n+1)(2n+1)}{6} + 3\cdot\frac{n(n+1)}{2} + n\Bigr] = \frac{1}{6}\Bigl[\frac{n(n+1)(2n+1)}{3} + \frac{3n(n+1)}{2} + n\Bigr] = \frac{n}{6}\Bigl[\frac{(n+1)(2n+1)}{3} + \frac{3(n+1)}{2} + 1\Bigr] = \frac{n}{36}\Bigl[2(n+1)(2n+1) + 9(n+1) + 6\Bigr] = \frac{n}{36}\bigl[4n^2 + 2n + 4n + 2 + 9n + 9 + 6\bigr] = \frac{n}{36}(4n^2 + 15n + 17). ]

Answer: ( \dfrac{n(4n^2 + 15n + 17)}{36} )