<p>If, for a positive integer \(n\), the quadratic equation \(x(x+1) + (x+1)(x+2) + \ldots + (x+\overline{n-1})(x+n) = 10n\) has two consecutive integral solutions, then \(n\) is equal to</p>
Step-by-Step Solution
Key Concept: Recognize that the left side is a sum of consecutive products of the form (x+k)(x+k+1). This can be rewritten using the telescoping identity (x+k)(x+k+1) = (x+k+1)² - (x+k+1), or equivalently, use the fact that consecutive integral solutions means if r is a solution, then r+1 is also a solution.
<p><strong>Step 1:</strong> Let the two consecutive integral solutions be r and r+1. The sum of roots is r + (r+1) = 2r+1 and the product is r(r+1).</p><p><strong>Step 2:</strong> Expand the left side. Each term (x+k)(x+k+1) expands to x² + (2k+1)x + k(k+1). Summing from k=0 to k=n-1:
∑(x+k)(x+k+1) = nx² + x∑(2k+1) + ∑k(k+1) = nx² + x·n² + [n(n+1)(n+2)/3 - n(n+1)/2]</p><p><strong>Step 3:</strong> Simplify: the constant term is n(n-1)(n+1)/3. The equation becomes: nx² + n²x + n(n²-1)/3 = 10n</p><p><strong>Step 4:</strong> Divide by n: x² + nx + (n²-1)/3 - 10 = 0, or x² + nx + (n²-31)/3 = 0</p><p><strong>Step 5:</strong> By Vieta's formulas: r(r+1) = (n²-31)/3. Also, 2r+1 = -n, so r = -(n+1)/2.</p><p><strong>Step 6:</strong> Substitute: [-(n+1)/2][-(n+1)/2 + 1] = (n²-31)/3 → [-(n+1)/2][(1-n)/2] = (n²-31)/3 → (n+1)(n-1)/4 = (n²-31)/3 → 3(n²-1) = 4(n²-31) → 3n² - 3 = 4n² - 124 → n² = 121 → n = 11</p><p>∴ Answer: A (n = 11)</p>
Correct Answer: A