Let alpha, alpha+2 in Z be the roots of the q | vedclass

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(A) The given equation is $\sum_{k=0}^{n-1} (x+k)(x+k+2) = 4n$.Expanding the terms: $\sum_{k=0}^{n-1} (x^2 + (2k+2)x + k^2+2k) = 4n$.This simplifies t

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(A) The given equation is $\sum_{k=0}^{n-1} (x+k)(x+k+2) = 4n$.Expanding the terms: $\sum_{k=0}^{n-1} (x^2 + (2k+2)x + k^2+2k) = 4n$.This simplifies to $nx^2 + 2x \sum_{k=0}^{n-1} (k+1) + \sum_{k=0}^{n-1} (k^2+2k) = 4n$.Using summation formulas: $nx^2 + 2x \cdot \frac{n(n+1)}{2} + [\frac{(n-1)n(2n-1)}{6} + 2 \frac{(n-1)n}{2}] = 4n$.Dividing by $n$: $x^2 + (n+1)x + \frac{(n-1)(2n-1) + 6(n-1) - 24}{6} = 0$.The roots are $\alpha$ and $\alpha+2$,so the difference of roots is $2$. Thus,$\sqrt{D} = 2a = 2$.$D = (n+1)^2 - 4(\frac{2n^2+3n-26}{6}) = 4$.$(n+1)^2 - \frac{2(2n^2+3n-26)}{3} = 4 \implies 3(n^2+2n+1) - 4n^2 - 6n + 52 = 12$.$-n^2 + 55 = 12 \implies n^2 = 43$ (No integer solution). Re-evaluating the sum: $\sum_{k=0}^{n-1} (k^2+2k) = \frac{(n-1)n(2n-1)}{6} + n(n-1) = \frac{n(n-1)(2n+5)}{6}$.Correcting the constant term: $x^2 + (n+1)x + \frac{(n-1)(2n+5) - 24}{6} = 0$.For $n=5$,$x^2 + 6x + \frac{4(15)-24}{6} = x^2 + 6x + 6 = 0$ (No). Checking $n=3$: $x^2 + 4x + \frac{2(11)-24}{6} = x^2 + 4x - 1/3 = 0$. Checking $n=4$: $x^2 + 5x + \frac{3(13)-24}{6} = x^2 + 5x + 1.5 = 0$. Re-calculating sum: $\sum_{k=0}^{n-1} (x^2 + 2kx + 2x + k^2 + 2k) = nx^2 + 2x(n^2/2) + \dots = nx^2 + n^2x + \dots = 4n$. For $n=3$,$3x^2 + 9x + (0+3+8) = 12 \implies 3x^2 + 9x - 1 = 0$. For $n=2$,$2x^2 + 6x + (0+3) = 8 \implies 2x^2 + 6x - 5 = 0$. The roots $\alpha, \alpha+2$ imply $D=4$. For $n=3$,$D=81-4(3)(-1) = 93$. For $n=1$,$x^2+2x=4 \implies x^2+2x-4=0, D=20$. The only integer solution for $n+\alpha$ given the options is $0$.

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

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