The positive integer \( n \), for which the solutions of the equation \( x(x+2) + (x+2)(x+4) + \dots + (x+2n-2)(x+2n) = \frac{8n}{3} \) are two consecutive even integers, is :
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For a quadratic equation \( ax^2 + bx + c = 0 \), the condition for roots to differ by \( k \) is \( D = k^2 a^2 \).
has solutions that are two consecutive even integers.
First, observe the pattern of terms in the left-hand sum: \( (x+2k)(x+2k+2) \) for \( k = 0, 1, 2, \ldots, n-1 \).
Each of these can be expanded as: \((x+2k)(x+2k+2) = (x+2k)^2 + 2(x+2k).\)
However, let's look for a simpler expression. Expand and simplify: \((x+2k)(x+2k+2) = x^2 + 4kx + 4k^2 + 2x + 4k = x^2 + (4k+2)x + (4k^2 + 4k).\)
This generates an arithmetic sequence for fixed \( x \) as \( n \)-th terms. The sum formula for an arithmetic sequence gives us: \(S = nx^2 + \left(2n(n-1)x + 4\frac{(n-1)n(2n-1)}{6} + 2n(n-1)\right)x.\)
On the right, simplify: \(\frac{8n}{3} = \frac{8n}{3}.\)
Given solutions are two consecutive even integers, let's define \( x = 2m \) and \( x+2 = 2m+2 \). Therefore, these values satisfy the equation: \(\left(S = \frac{8n}{3}\right)\)
Now check each option to determine when the sum equals correct \( x \).
For \( n = 3 \), test appropriate sums for even numbers gives: \((2)(2+2) + (2+2)(2+4) + (2+4)(2+6) = \frac{24}{3} = 8\)
The correct answer is thus: 3, as all terms satisfy the necessary condition for consecutive even integer solutions.
