The positive integer n, for which the solutions of the equation

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

are two consecutive even integers, is : [2026]

Question

The positive integer n, for which the solutions of the equation

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

are two consecutive even integers, is : [2026]

Smart Achievers · Accepted Answer

(1)

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

$\Rightarrow \sum_{r = 1}^{n} (x + 2 r - 2) (x + 2 r) = \frac{8 n}{3}$

$n x^{2} + 2 x \sum_{r = 1}^{n} (2 r - 1) + 4 \sum_{r = 1}^{n} r (r - 1) = \frac{8 n}{3}$

$n x^{2} + 2 x n^{2} + \frac{4 n (n^{2} - 1)}{3} - \frac{8 n}{3} = 0$

$x^{2} + 2 n x + \frac{4 (n^{2} - 1)}{3} - \frac{8}{3} = 0$

$∵$ $| α - β | = 2 \Rightarrow \frac{\sqrt{D}}{| a |} = 2 \Rightarrow D = 4$

$\Rightarrow 4 n^{2} - 4 (\frac{4 (n^{2} - 1)}{3} - \frac{8}{3}) = 4$

$\Rightarrow n^{2} - \frac{4 n^{2}}{3} = - 3$

$\Rightarrow n^{2} = 9$

$\Rightarrow n = 3$

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