JEE Main 2026 Mathematics question
Let α, α + 2, α ∈ ℤ, be the roots of the quadratic equation x(x + 2) + (x + 1)(x + 3) + (x + 2)(x + 4) + … + (x + n − 1)(x + n + 1) = 4n for some n ∈ ℕ. Then n + α is equal to:
Answer: (C)
Solution
Each term is x² + (2i + 2)x + i(i + 2), so the equation is nx² + n(n + 1)x + [(n − 1)n(2n − 1)/6 + n(n − 1)] = 4n. Dividing by n: x² + (n + 1)x + (n − 1)(2n + 5)/6 − 4
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