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JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations

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

  1. A \(0\)
  2. B \(1\)
  3. C \(2\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2\)

Step-by-step Solution

Detailed explanation

The given equation is: \(\sum_{k=0}^{n-1} (x+k)(x+k+2) = 4n\) Expanding the terms inside the summation: \(\sum_{k=0}^{n-1} (x^2 + 2(k+1)x + k(k+2)) = 4n\) Summing each term separately: \(n x^2 + 2x \sum_{k=0}^{n-1} (k+1) + \sum_{k=0}^{n-1} (k^2 + 2k) = 4n\) Using the standard…
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