Question

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 :

• A. 9
• B. 3
• C. 12
• D. 6

Hint:

For a quadratic equation \( ax^2 + bx + c = 0 \), the condition for roots to differ by \( k \) is \( D = k^2 a^2 \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We simplify the sum of quadratic terms to form a single quadratic equation. The condition "consecutive even integers" implies that the difference between the roots is 2.

Step 2: Detailed Explanation: The equation is \( \sum_{k=0}^{n-1} (x+2k)(x+2k+2) = \frac{8n}{3} \).
Expanding the general term: \( x^2 + (4k+2)x + 4k(k+1) \).
Summing from \( k = 0 \) to \( n-1 \):
\( n x^2 + [4 \frac{(n-1)n}{2} + 2n] x + 4 [\frac{(n-1)n(2n-1)}{6} + \frac{(n-1)n}{2}] = \frac{8n}{3} \).
Divide by \( n \):
\( x^2 + 2nx + \frac{2}{3}(n-1)(2n-1) + 2(n-1) - \frac{8}{3} = 0 \).
Simplifying the constant term:
\( x^2 + 2nx + \frac{4n^2 - 12}{3} = 0 \).
If roots \( \alpha, \beta \) are consecutive even integers, then \( |\alpha - \beta| = 2 \).
Using the relation \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \):
\[ 2^2 = (-2n)^2 - 4(\frac{4n^2 - 12}{3}) \]
\[ 4 = 4n^2 - \frac{16n^2 - 48}{3} \]
\[ 12 = 12n^2 - 16n^2 + 48 \implies 4n^2 = 36 \implies n = 3 \).

Step 3: Final Answer:
The positive integer \( n \) is 3.