Let $$\alpha, \alpha + 2, \alpha \in \mathbb{Z}$$, be the roots of the quadratic equation $$x(x+2) + (x+1)(x+3) + (x+2)(x+4) + \ldots + (x+n-1)(x+n+1) = 4n$$ for some $$n \in \mathbb{N}$$. Then $$n + \alpha$$ is equal to :

The given equation is:

$$x(x + 2) + (x + 1)(x + 3) + (x + 2)(x + 4) + \dots + (x + n - 1)(x + n + 1) = 4n$$

The $$r$$-th term of the series on the left-hand side can be written as:

$$T_r = (x + r - 1)(x + r + 1) = (x + r)^2 - 1^2$$

where $$r$$ goes from $$1$$ to $$n$$.

Summing $$T_r$$ from $$r=1$$ to $$n$$:

$$\sum_{r=1}^{n} [(x + r)^2 - 1] = 4n$$

$$\sum_{r=1}^{n} (x^2 + 2rx + r^2) - \sum_{r=1}^{n} 1 = 4n$$

Now apply the standard summation formulas:

• $$\sum_{r=1}^{n} x^2 = nx^2$$
• $$\sum_{r=1}^{n} 2rx = 2x \sum r = 2x \frac{n(n+1)}{2} = n(n+1)x$$
• $$\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$$
• $$\sum_{r=1}^{n} 1 = n$$

Substitute these back into the equation:

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

Divide the entire equation by $$n$$ (since $$n \in \mathbb{N}$$, $$n \neq 0$$):

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

$$x^2 + (n+1)x + \frac{2n^2 + 3n + 1 - 30}{6} = 0$$

$$x^2 + (n+1)x + \frac{2n^2 + 3n - 29}{6} = 0$$

We are told the roots are $$\alpha$$ and $$\alpha + 2$$.

• Sum of roots: $$\alpha + (\alpha + 2) = -(n+1)$$  $$2\alpha + 2 = -n - 1 \implies 2\alpha + n = -3 \quad \text{---(Eq. 1)}$$
• Difference of roots: $$(\alpha + 2) - \alpha = 2$$

  The difference of roots is also given by $$\frac{\sqrt{D}}{|A|}$$:

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

  Square both sides:

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

  Multiply by 3:

  $$3(n^2 + 2n + 1) - (4n^2 + 6n - 58) = 12$$ $$3n^2 + 6n + 3 - 4n^2 - 6n + 58 = 12$$ $$-n^2 + 61 = 12 \implies n^2 = 49 \implies n = 7$$

• (since $$n \in \mathbb{N}$$)

Substitute $$n = 7$$ into Eq. 1:

$$2\alpha + 7 = -3$$  $$2\alpha = -10 \implies \alpha = -5$$

Now, calculate $$n + \alpha$$:

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

The correct option is C (2).