Quadratic Equations
Sum of Consecutive Products
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>(a) 11</p>
<p>(b) 12</p>
<p>(c) 9</p>
<p>(d) 10</p>
Step-by-Step Solution
Key Concept: Sum the products using telescoping or polynomial expansion, then use Vieta's formulas with the constraint that roots are consecutive integers.
<p><strong>Solution:</strong> Expand the left side:</p><p>$\sum_{k=0}^{n-1}(x+k)(x+k+1) = \sum_{k=0}^{n-1}(x^2 + (2k+1)x + k(k+1))$</p><p>$= nx^2 + x\sum_{k=0}^{n-1}(2k+1) + \sum_{k=0}^{n-1}k(k+1)$</p><p>$= nx^2 + xn^2 + \frac{n(n-1)(n+1)}{3}$</p><p>The equation becomes: $nx^2 + n^2x + \frac{n(n-1)(n+1)}{3} = 10n$</p><p>Dividing by n: $x^2 + nx + \frac{(n-1)(n+1)}{3} - 10 = 0$</p><p>For two consecutive integral roots $m$ and $m+1$:</p><p>Sum = $2m + 1 = -n$, so $m = -\frac{n+1}{2}$</p><p>Product = $m(m+1) = \frac{(n-1)(n+1)}{3} - 10$</p><p>Substituting and solving: $n = 10$</p><p>∴ Answer is (d).</p>
Correct Answer: D