Let S_n = sum_{k=1}^{4n} (-1)^{frac{k(k+1)}{2 | vedclass

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(A,D) The given sum is $S_n = \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}} k^2$. Expanding the terms: $S_n = -1^2 - 2^2 + 3^2 + 4^2 - 5^2 - 6^2 + 7^2 + 8^2

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$(A, D)$

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(A,D) The given sum is $S_n = \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}} k^2$. Expanding the terms: $S_n = -1^2 - 2^2 + 3^2 + 4^2 - 5^2 - 6^2 + 7^2 + 8^2 + \dots + (-1)^{\frac{4n(4n+1)}{2}} (4n)^2$. Grouping the terms in sets of four: $S_n = (3^2 - 1^2) + (4^2 - 2^2) + (7^2 - 5^2) + (8^2 - 6^2) + \dots + ((4n-1)^2 - (4n-3)^2) + ((4n)^2 - (4n-2)^2)$. Using $a^2 - b^2 = (a-b)(a+b)$: $S_n = (2)(4) + (2)(6) + (2)(12) + (2)(14) + \dots + (2)(8n-4) + (2)(8n-2)$. $S_n = 2 [ (4 + 12 + \dots + 8n-4) + (6 + 14 + \dots + 8n-2) ]$. Both series inside the brackets have $n$ terms and are arithmetic progressions. Sum of first series $= \frac{n}{2} [4 + (8n-4)] = 4n^2$. Sum of second series $= \frac{n}{2} [6 + (8n-2)] = n(4n+2) = 4n^2 + 2n$. $S_n = 2 [ 4n^2 + 4n^2 + 2n ] = 2 [ 8n^2 + 2n ] = 4n(4n+1)$. For $n=8, S_8 = 4(8)(32+1) = 32 	imes 33 = 1056$. For $n=9, S_9 = 4(9)(36+1) = 36 	imes 37 = 1332$. Thus,$S_n$ can take values $1056$ and $1332$.

Let $S_n = \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}} k^2$. Then $S_n$ can take the value$(s)$:
$(A) 1056$
$(B) 1088$
$(C) 1120$
$(D) 1332$

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Let $S_n = \sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}} k^2$. Then $S_n$ can take the value$(s)$: $(A) 1056$ $(B) 1088$ $(C) 1120$ $(D) 1332$