<p>Find the sum \(11^2 - 1^2 + 12^2 - 2^2 + 13^2 - 3^2 + \cdots + 20^2 - 10^2\).</p>
Step-by-Step Solution
Key Concept: Group consecutive terms as differences of squares, then apply the factorization a² - b² = (a+b)(a-b) to convert each pair into a linear expression that sums easily.
<p><strong>Step 1:</strong> Group the terms in pairs:</p><p>(11² - 1²) + (12² - 2²) + (13² - 3²) + ⋯ + (20² - 10²)</p><p><strong>Step 2:</strong> Apply difference of squares: a² - b² = (a+b)(a-b)</p><p>(11+1)(11-1) + (12+2)(12-2) + (13+3)(13-3) + ⋯ + (20+10)(20-10)</p><p>= 12(10) + 14(10) + 16(10) + ⋯ + 30(10)</p><p><strong>Step 3:</strong> Factor out 10:</p><p>= 10(12 + 14 + 16 + ⋯ + 30)</p><p><strong>Step 4:</strong> Sum the arithmetic sequence 12, 14, 16, ..., 30 (10 terms with first term a=12, last term l=30):</p><p>Sum = (10/2)(12 + 30) = 5(42) = 210</p><p><strong>Step 5:</strong> Final answer:</p><p>10 × 210 = 1400</p><p>∴ Answer: <strong>1400</strong></p>
Correct Answer: 1400