Roots of Quadratic Equation Formed by Sum of Products

To find the value of $n + \alpha$, we first simplify the given quadratic equation.

The given equation is: $x(x+2) + (x+1)(x+3) + (x+2)(x+4) + \dots + (x+n-1)(x+n+1) = 4n$

The $k$-th term of the summation on the left-hand side, for $k = 1, 2, \dots, n$, can be written as: $T_k = (x + k - 1)(x + k + 1) = (x + k)^2 - 1 = x^2 + 2kx + k^2 - 1$

Now, summing $T_k$ from $k = 1$ to $n$: $\sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \left(x^2 + 2kx + k^2 - 1\right)$ $\sum_{k=1}^{n} T_k = n x^2 + 2x \sum_{k=1}^{n} k + \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} 1$

Using the standard formulas for the sum of natural numbers and their squares: $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$ $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

Substituting these back into the sum: $\sum_{k=1}^{n} T_k = n x^2 + 2x \cdot \frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6} - n$ $= n x^2 + n(n+1)x + \frac{n(n+1)(2n+1) - 6n}{6}$

Equating this to $4n$: $n x^2 + n(n+1)x + \frac{n(n+1)(2n+1) - 6n}{6} = 4n$

Since $n \in \mathbb{N}$, we can divide the entire equation by $n$: $x^2 + (n+1)x + \frac{(n+1)(2n+1) - 6}{6} = 4$ $x^2 + (n+1)x + \frac{2n^2 + 3n + 1 - 6 - 24}{6} = 0$ $x^2 + (n+1)x + \frac{2n^2 + 3n - 29}{6} = 0$

Let the roots of this quadratic equation be $x_1 = \alpha$ and $x_2 = \alpha + 2$. The absolute difference between the roots is: $|x_2 - x_1| = |(\alpha + 2) - \alpha| = 2$

For a quadratic equation $Ax^2 + Bx + C = 0$, the difference of roots is given by $\sqrt{B^2 - 4AC} / |A|$. Here $A = 1$, $B = n+1$, and $C = \frac{2n^2 + 3n - 29}{6}$.

Therefore: $\sqrt{(n+1)^2 - 4 \left(\frac{2n^2 + 3n - 29}{6}\right)} = 2$

Squaring both sides: $(n+1)^2 - \frac{2(2n^2 + 3n - 29)}{3} = 4$

Multiplying the entire equation by 3: $3(n^2 + 2n + 1) - 2(2n^2 + 3n - 29) = 12$ $3n^2 + 6n + 3 - 4n^2 - 6n + 58 = 12$ $-n^2 + 61 = 12$ $n^2 = 49$

Since $n \in \mathbb{N}$, we get: $n = 7$

Substitute $n = 7$ back into the simplified quadratic equation to find the roots: $x^2 + (7+1)x + \frac{2(7)^2 + 3(7) - 29}{6} = 0$ $x^2 + 8x + \frac{98 + 21 - 29}{6} = 0$ $x^2 + 8x + \frac{90}{6} = 0$ $x^2 + 8x + 15 = 0$

Factoring the quadratic equation gives: $(x + 3)(x + 5) = 0$

Thus, the roots are $x = -5$ and $x = -3$.

Comparing these roots with $\alpha$ and $\alpha + 2$: $\alpha = -5 \quad \text{and} \quad \alpha + 2 = -3$

Since $\alpha = -5 \in \mathbb{Z}$, this value is valid.

Finally, calculate $n + \alpha$: $n + \alpha = 7 + (-5) = 2$