Sequences & Series
Sum of Series
Grade 11
Question:
<p>If, for a positive integer \(n\), the quadratic equation, \(x(x+1) + (x+1)(x+2) + \cdots + (x+n-1)(x+n) = 10n\) has two consecutive integral solutions, then \(n\) is equal to</p>
<p>9</p>
<p>10</p>
<p>11</p>
<p>12</p>
Step-by-Step Solution
Key Concept: The LHS is a sum of n consecutive products of the form (x+k)(x+k+1). Use telescoping or expand to get a quadratic in x, then apply the condition that two consecutive integers satisfy it simultaneously to find n.
<p><strong>Step 1: Simplify the LHS using telescoping.</strong></p><p>Notice that (x+k)(x+k+1) = (x+k)² + (x+k). Sum from k=0 to n-1:</p><p>LHS = Σ[(x+k)² + (x+k)] = Σ(x+k)² + Σ(x+k)</p><p><strong>Step 2: Evaluate the sums.</strong></p><p>Σ(x+k)² from k=0 to n-1 gives n·x² + 2x·(0+1+...+(n-1)) + (0²+1²+...+(n-1)²)</p><p>= nx² + 2x·[n(n-1)/2] + [(n-1)n(2n-1)/6]</p><p>Σ(x+k) from k=0 to n-1 = nx + n(n-1)/2</p><p>LHS = nx² + n(n-1)x + (n-1)n(2n-1)/6 + nx + n(n-1)/2</p><p>= nx² + n(n)x + n(n-1)/2 + (n-1)n(2n-1)/6</p><p><strong>Step 3: Simplify to standard form.</strong></p><p>nx² + n²x + n(n-1)[3 + (2n-1)]/6 = 10n</p><p>Divide by n: x² + nx + (n-1)(2n+2)/6 = 10</p><p>x² + nx + (n-1)(n+1)/3 - 10 = 0</p><p><strong>Step 4: Apply condition for consecutive integer roots.</strong></p><p>If x = r and x = r+1 are roots, their sum is 2r+1 = -n (by Vieta's), so r = -(n+1)/2.</p><p>Product: r(r+1) = (n-1)(n+1)/3 - 10</p><p>For r = -(n+1)/2: [-(n+1)/2][-(n-1)/2] = (n²-1)/4</p><p>(n²-1)/4 = (n²-1)/3 - 10</p><p>(n²-1)[1/4 - 1/3] = -10 ⟹ (n²-1)(-1/12) = -10</p><p>n² - 1 = 120 ⟹ n² = 121 ⟹ n = 11</p><p><strong>∴ Answer: C (n = 11)</strong></p>
Correct Answer: C