The positive integer n, for which the solutio | vedclass

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(A) The given equation is $\sum_{r=1}^{n}(x+2r-2)(x+2r) = \frac{8n}{3}$.Expanding the terms: $\sum_{r=1}^{n}(x^2 + 4rx - 2x + 4r^2 - 4r) = \frac{8n}{3

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$3$

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(A) The given equation is $\sum_{r=1}^{n}(x+2r-2)(x+2r) = \frac{8n}{3}$.Expanding the terms: $\sum_{r=1}^{n}(x^2 + 4rx - 2x + 4r^2 - 4r) = \frac{8n}{3}$.This simplifies to $nx^2 + 2x(2\sum r - n) + 4\sum r^2 - 4\sum r = \frac{8n}{3}$.Using $\sum_{r=1}^{n} r = \frac{n(n+1)}{2}$ and $\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$, we get $nx^2 + 2x(n^2) + \frac{4n(n+1)(2n+1)}{6} - 2n(n+1) = \frac{8n}{3}$.Dividing by $n$: $x^2 + 2nx + \frac{2(n+1)(2n+1)}{3} - 2(n+1) = \frac{8}{3}$.$x^2 + 2nx + \frac{4n^2+6n+2-6n-6-8}{3} = 0 \Rightarrow x^2 + 2nx + \frac{4n^2-12}{3} = 0$.Since the roots $\alpha, \beta$ are consecutive even integers, $|\alpha - \beta| = 2$.Thus, $\frac{\sqrt{D}}{1} = 2 \Rightarrow D = 4$.$D = (2n)^2 - 4(\frac{4n^2-12}{3}) = 4$.$4n^2 - \frac{16n^2-48}{3} = 4 \Rightarrow 12n^2 - 16n^2 + 48 = 12$.$-4n^2 = -36$ $\Rightarrow n^2 = 9$ $\Rightarrow n = 3$.

The positive integer $n$, for which the solutions of the equation $x(x+2)+(x+2)(x+4)+...+(x+2n-2)(x+2n) = \frac{8n}{3}$ are two consecutive even integers, is:

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