The sign \((-1)^{K(K+1)/2}\) follows the repeating pattern \(-,-,+,+\) for \(K = 1,2,3,4\) and then repeats with period 4, since \(K(K+1)/2 \pmod 2\) cycles through odd, odd, even, even values.

Group the terms in blocks of 4 consecutive integers starting at \(4m+1\): each block contributes \(-[(4m{+}1)^2+(4m{+}2)^2]+[(4m{+}3)^2+(4m{+}4)^2]\).

Using the difference of squares, \((4m{+}3)^2-(4m{+}1)^2 = 16m+8\) and \((4m{+}4)^2-(4m{+}2)^2 = 16m+12\), so each block sums to \(32m+20\).

Summing over \(m = 0\) to \(n-1\): \(S_n = \sum_{m=0}^{n-1}(32m+20) = 16n(n-1)+20n = 16n^2+4n = 4n(4n+1)\).

Check for \(n=1\): direct computation gives \(-1-4+9+16=20\), and the formula gives \(4(1)(5)=20\), confirming the formula.

For \(n=8\): \(S_8 = 4(8)(33) = 1056\), which matches the given value 1056.