Question

Let \( \alpha, \alpha + 2 \in \mathbb{Z} \) be the roots of the quadratic equation} \[ x(x+2) + (x+1)(x+3) + (x+2)(x+4) + \cdots + (x+n-1)(x+n+1) = 4n \] for some \( n \in \mathbb{N} \). Then \( n + \alpha \) is equal to:

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Answer 1

The Correct Option is C

Concept: The expression given is a sum of quadratic terms. Each term follows a pattern: \[ (x+k)(x+k+2) \] Key algebraic ideas used:

• Expansion of quadratic expressions
• Sum of natural numbers: \( \sum_{k=0}^{n-1} k = \frac{n(n-1)}{2} \)
• Sum of squares: \( \sum_{k=0}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6} \)
• Relations between roots and coefficients of a quadratic equation

Step 1:Express the general term of the sum.} The general term is \[ (x+k)(x+k+2) \] Expanding, \[ (x+k)(x+k+2) = x^2 + 2x + 2kx + k^2 + 2k \] Now summing from \(k=0\) to \(n-1\): \[ \sum (x+k)(x+k+2) \] \[ = n x^2 + 2n x + 2x\sum k + \sum k^2 + 2\sum k \] Using the standard summation formulas, \[ \sum k = \frac{n(n-1)}{2}, \qquad \sum k^2 = \frac{(n-1)n(2n-1)}{6} \] Substituting, \[ S = n x^2 + 2n x + x n(n-1) + \frac{(n-1)n(2n-1)}{6} + n(n-1) \] Given that \[ S = 4n \]
Step 2:Form the quadratic equation in \(x\).} \[ n x^2 + n(n+1)x + \frac{(n-1)n(2n-1)}{6} + n(n-1) - 4n = 0 \] Dividing by \(n\): \[ x^2 + (n+1)x + \frac{(n-1)(2n+5)-24}{6} = 0 \]
Step 3:Use the given roots.} Roots are \( \alpha \) and \( \alpha + 2 \). Using sum of roots: \[ \alpha + (\alpha+2) = -(n+1) \] \[ 2\alpha + 2 = -(n+1) \] \[ \alpha = -\frac{n+3}{2} \]
Step 4:Use product of roots.} \[ \alpha(\alpha+2) = \frac{(n+3)(n-1)}{4} \] But product of roots is also \[ \frac{(n-1)(2n+5)-24}{6} \] Equating, \[ \frac{(n+3)(n-1)}{4} = \frac{(n-1)(2n+5)-24}{6} \] Multiplying by \(12\): \[ 3(n+3)(n-1) = 2[(n-1)(2n+5)-24] \] Solving, \[ n^2 - 49 = 0 \] \[ n = 7 \]
Step 5:Find \( \alpha \).} \[ \alpha = -\frac{n+3}{2} = -\frac{10}{2} = -5 \] Therefore, \[ n + \alpha = 7 - 5 = 2 \]