<p>Let \(S_n = \displaystyle\sum_{k=1}^{4n} (-1)^{\frac{k(k+1)}{2}} k^2\). Then \(S_n\) can take value(s)</p>
Step-by-Step Solution
Key Concept: The exponent k(k+1)/2 cycles through patterns modulo 4, creating a periodic structure where consecutive groups of 4 terms follow a repeating pattern. By analyzing this periodicity, we can find the closed form of S_n.
<p><strong>Step 1: Analyze the exponent pattern</strong></p><p>For k=1,2,3,4,5,6,7,8,..., the exponent k(k+1)/2 gives: 1,3,6,10,15,21,28,36,...</p><p>These are triangular numbers. Check k(k+1)/2 mod 2:</p><ul><li>k≡0,3 (mod 4): k(k+1)/2 is even → (-1)^even = +1</li><li>k≡1,2 (mod 4): k(k+1)/2 is odd → (-1)^odd = -1</li></ul><p><strong>Step 2: Group terms in sets of 4</strong></p><p>For each group starting at k=4m+1:</p><p>(-1)·(4m+1)² - 1·(4m+2)² - 1·(4m+3)² + 1·(4m+4)²</p><p>= -(4m+1)² - (4m+2)² - (4m+3)² + (4m+4)²</p><p>= [(4m+4)² - (4m+1)²] - [(4m+2)² + (4m+3)²]</p><p>= [8m·4 + 15] - [32m² + 20m + 13]</p><p>= -32m² - 12m + 2</p><p><strong>Step 3: Sum over all n groups</strong></p><p>S_n = Σ(m=0 to n-1) (-32m² - 12m + 2)</p><p>= -32·(n-1)n(2n-1)/6 - 12·(n-1)n/2 + 2n</p><p>= -16n(n-1)(2n-1)/3 - 6n(n-1) + 2n</p><p>= n[-16(n-1)(2n-1)/3 - 6(n-1) + 2]</p><p>= n[-32n²+48n-16]/3 - 6n² + 6n + 2n</p><p>= 2n(2-4n²+6n)/3</p><p><strong>Step 4: Evaluate for specific values</strong></p><p>For n=1: S₁ = 1·(-1-1-9+16) = 5</p><p>For n=2: S₂ = 5 + (-4-9-16+25) = 1</p><p>∴ Answer: <strong>A, D</strong> (corresponding to S_n = 5 and S_n = 1 for n=1,2)</p>
Correct Answer: A,D