Question:

<p><strong>Step 1:</strong> Group consecutive pairs of squares using difference of squares formula.</p><p>S_n = (1² - 2²) + (3² - 4²) + (5² - 6²) + ...</p><p><strong>Step 2:</strong> Apply a² - b² = (a+b)(a-b) to each pair.</p><p>(1² - 2²) = (1+2)(1-2) = -3</p><p>(3² - 4²) = (3+4)(3-4) = -7</p><p>(5² - 6²) = (5+6)(5-6) = -11</p><p><strong>Step 3:</strong> For <strong>even n = 2m</strong>: We have m complete pairs.</p><p>S_{2m} = -3 - 7 - 11 - ... - (4m-1)</p><p>This is an AP with first term a = -3, last term l = -(4m-1), and m terms.</p><p>S_{2m} = -m(3 + 4m - 1)/2 = -m(4m + 2)/2 = <strong>-m(2m+1)</strong></p><p><strong>Step 4:</strong> For <strong>odd n = 2m+1</strong>: We have m complete pairs plus the single term (2m+1)².</p><p>S_{2m+1} = S_{2m} + (2m+1)² = -m(2m+1) + (2m+1)²</p><p>= (2m+1)[-m + 2m + 1] = <strong>(2m+1)(m+1)</strong></p><p><strong>Step 5:</strong> Express in terms of n directly:</p><p>• If n is even (n = 2m): S_n = <strong>-n(n+2)/4</strong></p><p>• If n is odd (n = 2m+1): S_n = <strong>n(n+1)/4</strong></p><p>Or combined: S_n = <strong>(-1)^{n+1}·n⌈n/2⌉</strong></p><p>∴ Answer: A,C,D (statements involving these formulas)</p>