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

(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) )