Question

$x(x + 2) + (x + 2)(x + 4) + \ldots. + (x + 2n - 2)(x + 2n) = \frac{8n}{3}$

$${\Rightarrow \sum_{r = 1}^{n}\mspace{2mu}(x + 2r - 2)(x + 2r) = \frac{8n}{3} }{nx^{2} + 2x\sum_{r = 1}^{n}\mspace{2mu}(2r - 1) + 4\sum_{r = 1}^{n}\mspace{2mu} r(r - 1) = \frac{8n}{3} }{nx^{2} + 2x \cdot n^{2} + \frac{4n\left( n^{2} - 1 \right)}{3} - \frac{8n}{3} = 0 }{x^{2} + 2nx + \frac{4\left( n^{2} - 1 \right)}{3} - \frac{8}{3} = 0 <_{\beta}^{\alpha} }{\because|\alpha - \beta| = 2 \Rightarrow \frac{\sqrt{D}}{|a|} = 2 \Rightarrow D = 4 }{\Rightarrow 4n^{2} - 4\left( 4\frac{\left( n^{2} - 1 \right)}{3} - \frac{8}{3} \right) = 4 }{\Rightarrow n^{2} - \frac{4n^{2}}{3} = - 3 }{\Rightarrow n^{2} = 9 }{\Rightarrow n = 3}$$