Let $S_n = \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}} k^2$. Then $S_n$ can take value(s)

Let Sₙ = ∑ₖ₌₁⁴ⁿ (-1)^((k(k+1))/2) k². Then Sₙ can take value(s)

$\text{Given: } S_n = \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}}\, k^2$

$\text{To find: which option(s) can equal } S_n$

$(1)\ 1056 \quad (2)\ 1088 \quad (3)\ 1120 \quad (4)\ 1332$

$T_k = \dfrac{k(k+1)}{2}$

$k=1,2,3,4:\quad T_k = 1,3,6,10 \;\Rightarrow\; \text{signs } -,-,+,+$

$k=5,6,7,8:\quad T_k = 15,21,28,36 \;\Rightarrow\; \text{signs } -,-,+,+$

$\text{sign pattern period } = 4$

$1.\ \text{group terms in blocks of } 4$

$2.\ \text{find each block's sum}$

$3.\ \text{add blocks} \Rightarrow S_n,\ \text{then match with options}$

$\text{block } m:\ k=4m+1,\,4m+2,\,4m+3,\,4m+4$

$m=0,1,\dots,n-1 \quad (\,n\text{ blocks cover }4n\text{ terms}\,)$

$B_m = -(4m+1)^2 - (4m+2)^2 + (4m+3)^2 + (4m+4)^2$

$(4m+3)^2-(4m+1)^2 = [\,(4m+3)-(4m+1)\,][\,(4m+3)+(4m+1)\,]$

$= 2\,(8m+4) = 16m+8$

$(4m+4)^2-(4m+2)^2 = 2\,(8m+6) = 16m+12$

$B_m = (16m+8)+(16m+12) = 32m+20$

$B_0=20,\ B_1=52,\ B_2=84,\dots \quad (d=32)$

$S_n = \sum_{m=0}^{n-1}(32m+20)$

$= 32\sum_{m=0}^{n-1} m \;+\; 20\sum_{m=0}^{n-1} 1$

$\sum_{m=0}^{n-1} m = \dfrac{n(n-1)}{2}, \quad \sum_{m=0}^{n-1} 1 = n$

$S_n = 32\cdot\dfrac{n(n-1)}{2}+20n = 16n(n-1)+20n$

$S_n = 16n^2+4n = 4n(4n+1)$

$S_n = 4n(4n+1) \;\Rightarrow\; 4 \mid S_n$

$\dfrac{S_n}{4} = n(4n+1)$

$1056 \div 4 = 264$

$n(4n+1) = 8(33) = 264 \;\Rightarrow\; n=8 \quad\checkmark$

$1332 \div 4 = 333$

$n(4n+1) = 9(37) = 333 \;\Rightarrow\; n=9 \quad\checkmark$

$1088 \div 4 = 272$

$n=8:\ 264,\ n=9:\ 333 \;\Rightarrow\; n(4n+1)\neq 272 \;\Rightarrow\; \text{reject}$

$1120 \div 4 = 280$

$n=8:\ 264,\ n=9:\ 333 \;\Rightarrow\; n(4n+1)\neq 280 \;\Rightarrow\; \text{reject}$

$\boxed{S_n = 1056\ (n=8) \quad \text{or} \quad S_n = 1332\ (n=9)}$

$\text{Answer: options (1) and (4)}$