Let $\alpha$, $\alpha + 2$, $\alpha \in Z$, 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 $n \in N$. Then $n + \alpha$ is equal to :
- (1) 0
- (2) 1
- (3) 2
- (4) 3
Show answer and solution
Answer: Option (3)
Key idea: Quadratic Equations with Series Summation and Vieta's Formulas
- Expand the general term: $(x+k)(x+k+2) = x^2 + (2k+2)x + k^2 + 2k$ for $k = 0$ to $n-1$
- Sum all terms using summation formulas: $\sum_{k=0}^{n-1}k = \frac{n(n-1)}{2}$ and $\sum_{k=0}^{n-1}k^2 = \frac{n(n-1)(2n-1)}{6}$
- Simplify to get: $x^2 + (n+1)x + \frac{2n^2 + 3n - 29}{6} = 0$
- Apply Vieta's formulas: Sum of roots $2\alpha + 2 = -(n+1)$ gives $\alpha = \frac{-n-3}{2}$
- Use product of roots: $\alpha(\alpha+2) = \frac{2n^2 + 3n - 29}{6}$ and substitute $\alpha$ to get $n^2 = 49$, so $n = 7$
- Calculate $\alpha = \frac{-7-3}{2} = -5$, therefore $n + \alpha = 7 + (-5) = 2$