Let α, α + 2, α ∈ ℤ , be the roots of the quadratic equation x(x+2)...

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 summation formulas:

$n x^2 + 2x \dfrac{n(n+1)}{2} + \dfrac{(n-1)n(2n-1)}{6} + 2 \dfrac{(n-1)n}{2} = 4n$

$n x^2 + n(n+1)x + \dfrac{n(n-1)(2n+5)}{6} = 4n$

Since $n \in \mathbb{N}$, dividing the entire equation by $n$ gives:

$x^2 + (n+1)x + \dfrac{(n-1)(2n+5)}{6} - 4 = 0$

The roots of this quadratic equation are given as $\alpha$ and $\alpha + 2$.

Using the sum of the roots:

$\alpha + (\alpha + 2) = -(n+1)$

$2\alpha + 2 = -n - 1 \Rightarrow n = -2\alpha - 3$

Using the product of the roots:

$\alpha(\alpha + 2) = \dfrac{(n-1)(2n+5)}{6} - 4$

Substituting $n = -2\alpha - 3$ into the product equation:

$\alpha^2 + 2\alpha = \dfrac{(-2\alpha - 4)(-4\alpha - 1)}{6} - 4$

$6(\alpha^2 + 2\alpha) = 8\alpha^2 + 18\alpha + 4 - 24$

$6\alpha^2 + 12\alpha = 8\alpha^2 + 18\alpha - 20$

$2\alpha^2 + 6\alpha - 20 = 0$

$\alpha^2 + 3\alpha - 10 = 0$

$(\alpha + 5)(\alpha - 2) = 0$

Thus, $\alpha = -5$ or $\alpha = 2$.

If $\alpha = 2$, then $n = -2(2) - 3 = -7$, which is rejected since $n \in \mathbb{N}$.

If $\alpha = -5$, then $n = -2(-5) - 3 = 7$, which is valid.

Therefore, $n + \alpha = 7 + (-5) = 2$.

Answer: $2$