Step 1: Understanding the Concept:
The given equation is a summation of \( n \) terms, each being a product of two terms differing by 2.
We need to simplify this summation into a standard quadratic form \( Ax^2 + Bx + C = 0 \). Step 2: Key Formula or Approach:
The \( r \)-th term is \( T_r = (x + r - 1)(x + r + 1) = x^2 + 2x(r - 1) + (r - 1)(r + 1) + 2x \dots \)
Actually, a simpler form for the \( r \)-th term starting from \( r=1 \) to \( n \) is \( T_r = (x + r - 1)(x + r + 1) = (x + r)^2 - 1 \).
Let's use \( T_r = (x + r - 1)(x + r + 1) = x^2 + 2x(r) + r^2 - 1 \) where indices are shifted. Step 3: Detailed Explanation:
The given equation is \( \sum_{r=0}^{n-1} (x+r)(x+r+2) = 4n \).
General term: \( (x+r)(x+r+2) = x^2 + 2(r+1)x + r(r+2) \).
Summing over \( r = 0 \) to \( n-1 \):
\[ \sum_{r=0}^{n-1} [x^2 + 2x(r+1) + r^2 + 2r] = 4n \]
\[ nx^2 + 2x \left[ \frac{n(n+1)}{2} \right] + \sum_{r=0}^{n-1} (r^2 + 2r) = 4n \]
Using \( \sum r^2 = \frac{(n-1)n(2n-1)}{6} \) and \( \sum r = \frac{(n-1)n}{2} \):
Dividing the whole equation by \( n \):
\[ x^2 + (n+1)x + \frac{(n-1)(2n-1)}{6} + (n-1) - 4 = 0 \]
Since roots are \( \alpha \) and \( \alpha+2 \), the difference of roots is \( 2 \).
Using \( (\text{Diff})^2 = (\text{Sum})^2 - 4(\text{Prod}) \):
\[ 2^2 = (n+1)^2 - 4 \left( \frac{2n^2 - 3n + 1}{6} + n - 5 \right) \]
Solving for \( n \), we get \( n = 7 \).
Substituting \( n=7 \), the equation becomes \( x^2 + 8x + 15 = 0 \).
Roots are \( -3, -5 \). Here \( \alpha = -5 \).
\( \alpha + n = -5 + 7 = 2 \). Step 4: Final Answer:
The value of \( \alpha + n \) is 2.
