<p>The quadratic equation \(x(x+1) + (x+1)(x+2) + \cdots + [x+(n-1)](x+n) = 10n\) has two consecutive integral solutions. Find the value of \(n\).</p>
Step-by-Step Solution
Key Concept: Recognize that the sum of n products of consecutive terms can be simplified using telescoping or the identity (x+k)(x+k+1) = x² + (2k+1)x + k(k+1). The resulting quadratic equation will have integer solutions only for specific values of n.
<p><strong>Step 1:</strong> Expand the left side by summing from k=0 to n-1:</p><p>LHS = Σ(x+k)(x+k+1) = Σ[x² + (2k+1)x + k(k+1)]</p><p>= nx² + x·Σ(2k+1) + Σk(k+1)</p><p><strong>Step 2:</strong> Calculate the sums:</p><p>Σ(2k+1) from k=0 to n-1 = 2·[n(n-1)/2] + n = n² (sum of first n odd numbers)</p><p>Σk(k+1) from k=0 to n-1 = Σk² + Σk = [n(n-1)(2n-1)/6] + [n(n-1)/2] = n(n-1)(n+1)/3</p><p><strong>Step 3:</strong> The equation becomes:</p><p>nx² + n²x + n(n-1)(n+1)/3 = 10n</p><p>Divide by n: x² + nx + (n-1)(n+1)/3 = 10</p><p>x² + nx + (n²-1)/3 - 10 = 0</p><p>x² + nx + (n²-31)/3 = 0</p><p><strong>Step 4:</strong> For two consecutive integer roots r and r+1:</p><p>Sum of roots: r + (r+1) = 2r+1 = -n, so r = -(n+1)/2</p><p>Product of roots: r(r+1) = (n²-31)/3</p><p><strong>Step 5:</strong> Substitute r = -(n+1)/2:</p><p>[-(n+1)/2][-(n+1)/2 + 1] = (n²-31)/3</p><p>[-(n+1)/2][(1-n)/2] = (n²-31)/3</p><p>(n+1)(n-1)/4 = (n²-31)/3</p><p>(n²-1)/4 = (n²-31)/3</p><p>3(n²-1) = 4(n²-31)</p><p>3n² - 3 = 4n² - 124</p><p>121 = n²</p><p>n = 11 (taking positive value)</p><p><strong>Verification:</strong> For n=11: x² + 11x + 10 = 0 gives (x+1)(x+10) = 0, roots are -10 and -1 (consecutive integers) ✓</p><p>∴ Answer: <strong>11</strong></p>
Correct Answer: 11