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

Step 1: Understanding the Concept:
The given equation involves a sum of terms of the form \((x+k-1)(x+k+1)\).
Since \((a-1)(a+1) = a^2 - 1\), each term can be simplified to a square minus one.
The resulting sum will form a quadratic equation in \(x\).
The roots are given as \(\alpha\) and \(\alpha+2\), allowing us to use the properties of quadratic equations like the sum and product of roots or the discriminant.
Step 2: Key Formula or Approach:
1. General term: \(T_k = (x+k-1)(x+k+1) = (x+k)^2 - 1\).
2. Sum of series: \(\sum_{k=1}^{n} ((x+k)^2 - 1) = 4n\).
3. Roots of \(ax^2 + bx + c = 0\) are \(\alpha_1, \alpha_2\), then \(|\alpha_1 - \alpha_2| = \frac{\sqrt{D}}{|a|}\).
Step 3: Detailed Explanation:
Expand the summation:
\[ \sum_{k=1}^{n} (x^2 + 2kx + k^2 - 1) = 4n \]
\[ nx^2 + 2x \left(\sum_{k=1}^{n} k\right) + \sum_{k=1}^{n} k^2 - n = 4n \]
Substitute the standard summation formulas:
\[ nx^2 + 2x \frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6} - n = 4n \]
Divide by \(n\) (since \(n \in \mathbb{N}\)):
\[ x^2 + (n+1)x + \frac{(n+1)(2n+1)}{6} - 1 = 4 \]
\[ x^2 + (n+1)x + \left( \frac{2n^2 + 3n + 1}{6} - 5 \right) = 0 \]
Comparing with \(ax^2 + bx + c = 0\), we have \(a=1, b=n+1\).
The roots are \(\alpha\) and \(\alpha+2\). The difference of roots is \(|(\alpha+2) - \alpha| = 2\).
Using \(D = b^2 - 4ac = a^2(\text{difference of roots})^2\):
\[ (n+1)^2 - 4(1) \left( \frac{2n^2 + 3n + 1}{6} - 5 \right) = 2^2 \]
\[ (n+1)^2 - \frac{2(2n^2 + 3n + 1)}{3} + 20 = 4 \]
Multiply by 3 to clear the fraction:
\[ 3(n^2 + 2n + 1) - 4n^2 - 6n - 2 + 60 = 12 \]
\[ 3n^2 + 6n + 3 - 4n^2 - 6n - 2 + 60 - 12 = 0 \]
\[ -n^2 + 49 = 0 \Rightarrow n^2 = 49 \Rightarrow n = 7 \]
Now, use the sum of roots formula:
\[ \alpha + (\alpha + 2) = -(n+1) \]
\[ 2\alpha + 2 = -(7+1) = -8 \]
\[ 2\alpha = -10 \Rightarrow \alpha = -5 \]
The value of \(n + \alpha = 7 + (-5) = 2\).
Step 4: Final Answer:
The required value of \(n + \alpha\) is 2.