$x(x+2) + (x+1)(x+3) + (x+2)(x+4) + \dots + (x+n-1)(x+n+1) = 4n$ for some $n \in \mathbf{N}$.
Then $n+\alpha$ is equal to :
Quadratic Equation Derivation
The given equation involves a sum of terms: $x(x+2) + (x+1)(x+3) + (x+2)(x+4) + \dots + (x+n-1)(x+n+1) = 4n$ This sum can be written as $\sum_{k=0}^{n-1} (x+k)(x+k+2)$. Expanding the general term: $(x+k)(x+k+2) = (x+k)^2 + 2(x+k) = x^2 + (2k+2)x + (k^2+2k)$. Summing this expression from $k=0$ to $n-1$: $\sum_{k=0}^{n-1} [x^2 + (2k+2)x + (k^2+2k)] = nx^2 + x \sum_{k=0}^{n-1} (2k+2) + \sum_{k=0}^{n-1} (k^2+2k)$. Calculating the sums: $\sum_{k=0}^{n-1} (2k+2) = 2 \frac{(n-1)n}{2} + 2n = n(n-1) + 2n = n^2+n = n(n+1)$. $\sum_{k=0}^{n-1} (k^2+2k) = \frac{(n-1)n(2n-1)}{6} + 2 \frac{(n-1)n}{2} = \frac{n(n-1)(2n+5)}{6}$. Substituting these back into the equation: $nx^2 + n(n+1)x + \frac{n(n-1)(2n+5)}{6} = 4n$. Since $n \in \mathbf{N}$, we can divide by $n$: $x^2 + (n+1)x + \frac{(n-1)(2n+5)}{6} = 4$. Rearranging into standard quadratic form $ax^2+bx+c=0$: $x^2 + (n+1)x + \left(\frac{(n-1)(2n+5)}{6} - 4\right) = 0$.
Roots Analysis using Vieta's Formulas
Let the roots of this quadratic equation be $r_1$ and $r_2$. The question implies the roots are $\alpha$ and $\alpha+2$. Using Vieta's formulas: Sum of roots: $r_1 + r_2 = \alpha + (\alpha+2) = 2\alpha+2$. From the equation, the sum is $-(n+1)$. So, $2\alpha+2 = -(n+1)$ (Equation 1) Product of roots: $r_1 r_2 = \alpha(\alpha+2) = \alpha^2+2\alpha$. From the equation, the product is $\frac{(n-1)(2n+5)}{6} - 4$. So, $\alpha^2+2\alpha = \frac{(n-1)(2n+5)}{6} - 4$ (Equation 2)
Parameter Solving for n and alpha
From Equation 1: $2(\alpha+1) = -(n+1)$. This gives $\alpha+1 = -\frac{n+1}{2}$, so $\alpha = -\frac{n+1}{2} - 1 = -\frac{n+3}{2}$. Substitute this expression for $\alpha$ into Equation 2: $\left(-\frac{n+3}{2}\right)^2 + 2\left(-\frac{n+3}{2}\right) = \frac{(n-1)(2n+5)}{6} - 4$. $\frac{(n+3)^2}{4} - (n+3) = \frac{2n^2+3n-5}{6} - 4$. Simplify both sides: $\frac{n^2+6n+9 - 4(n+3)}{4} = \frac{2n^2+3n-5 - 24}{6}$. $\frac{n^2+2n-3}{4} = \frac{2n^2+3n-29}{6}$. Cross-multiplying: $6(n^2+2n-3) = 4(2n^2+3n-29)$. $6n^2+12n-18 = 8n^2+12n-116$. $2n^2 = 98$. $n^2 = 49$. Since $n \in \mathbf{N}$, $n=7$. Now find $\alpha$ using $n=7$: $\alpha = -\frac{n+3}{2} = -\frac{7+3}{2} = -\frac{10}{2} = -5$. The condition $\alpha \in \mathbf{Z}$ is satisfied as $-5$ is an integer.
Calculation of n + alpha
The question asks for the value of $n + \alpha$. $n + \alpha = 7 + (-5) = 2$.
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