Possible Values for Alternating Series of Squares

To determine the possible values that the sum $S_n = \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}} k^2$ can take, we first analyze the sign pattern of $(-1)^{\frac{k(k+1)}{2}}$ for values of $k$:

• For $k = 4m + 1$: $\frac{k(k+1)}{2} = \frac{(4m+1)(4m+2)}{2} = (4m+1)(2m+1) \implies \text{odd} \implies (-1)^{\frac{k(k+1)}{2}} = -1$

• For $k = 4m + 2$: $\frac{k(k+1)}{2} = \frac{(4m+2)(4m+3)}{2} = (2m+1)(4m+3) \implies \text{odd} \implies (-1)^{\frac{k(k+1)}{2}} = -1$

• For $k = 4m + 3$: $\frac{k(k+1)}{2} = \frac{(4m+3)(4m+4)}{2} = (4m+3)(2m+2) \implies \text{even} \implies (-1)^{\frac{k(k+1)}{2}} = +1$

• For $k = 4m + 4$: $\frac{k(k+1)}{2} = \frac{(4m+4)(4m+5)}{2} = (2m+2)(4m+5) \implies \text{even} \implies (-1)^{\frac{k(k+1)}{2}} = +1$

Thus, the sign pattern repeats every 4 terms as $-1, -1, +1, +1$.

We can group the terms of the sum into $n$ blocks, each containing 4 consecutive terms for $m = 0, 1, 2, \dots, n-1$: $T_m = -(4m+1)^2 - (4m+2)^2 + (4m+3)^2 + (4m+4)^2$

Simplifying $T_m$: $T_m = \left[(4m+3)^2 - (4m+1)^2\right] + \left[(4m+4)^2 - (4m+2)^2\right]$ $T_m = (4m+3 - 4m-1)(4m+3 + 4m+1) + (4m+4 - 4m-2)(4m+4 + 4m+2)$ $T_m = 2(8m+4) + 2(8m+6) = 16m + 8 + 16m + 12 = 32m + 20$

Now, summing $T_m$ from $m = 0$ to $n-1$: $S_n = \sum_{m=0}^{n-1} (32m + 20)$ $S_n = 32 \sum_{m=0}^{n-1} m + \sum_{m=0}^{n-1} 20$ $S_n = 32 \cdot \frac{(n-1)n}{2} + 20n$ $S_n = 16n(n-1) + 20n = 16n^2 + 4n = 4n(4n+1)$

Now we test the given options to check for integer values of $n \ge 1$:

1. Option (A): 1056 $4n(4n+1) = 1056 \implies 4n^2 + n - 264 = 0$ The discriminant is $\Delta = 1^2 - 4(4)(-264) = 1 + 4224 = 4225 = 65^2$. $n = \frac{-1 + 65}{8} = 8$ Since $n = 8$ is a positive integer, $S_n$ can equal $1056$.

2. Option (B): 1088 $4n(4n+1) = 1088 \implies 4n^2 + n - 272 = 0$ The discriminant is $\Delta = 1^2 - 4(4)(-272) = 1 + 4352 = 4353$, which is not a perfect square. Thus, no integer solution exists.

3. Option (C): 1120 $4n(4n+1) = 1120 \implies 4n^2 + n - 280 = 0$ The discriminant is $\Delta = 1^2 - 4(4)(-280) = 1 + 4480 = 4481$, which is not a perfect square. Thus, no integer solution exists.

4. Option (D): 1332 $4n(4n+1) = 1332 \implies 4n^2 + n - 333 = 0$ The discriminant is $\Delta = 1^2 - 4(4)(-333) = 1 + 5328 = 5329 = 73^2$. $n = \frac{-1 + 73}{8} = 9$ Since $n = 9$ is a positive integer, $S_n$ can equal $1332$.

Therefore, the correct options are A and D.