Question
Let $\alpha,\alpha + 2,\alpha \in \mathbf{Z}$, be the roots of the quadratic equation
$x(x + 2) + (x + 1)(x + 3) + (x + 2)(x + 4) + \ldots.. + (x + n - 1)(x + n + 1) = 4n$ for some $n \in N$.
Then $(n + \alpha)$ is equal to :
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Solution
Sol. [Key Concept: Expand the telescoping sum into a quadratic in $x$; its discriminant must be a perfect square, which fixes $n$.]
$${nx}^{2} + x(2 + 4 + 6 + \ldots.. + 2n) + \left( 1.3 + \ldots.. + (n - 1)(n + 1) \right) = 4n $$
$$nx^{2} + n(n + 1)x + \frac{n(n - 1)(2n + 5)}{6} = 4n $$
$$x^{2} + (n + 1)x + \frac{(n - 1)(2n + 5)}{6} = 4 $$D must be a perfect square
$$D = \frac{122 - 2n^{2}}{6} = 20 - \left( \frac{n^{2} - 1}{3} \right) $$If make perfect square $\Rightarrow \frac{n^{2} - 1}{3} = 16 \Rightarrow n = 7$
So Equation is
$$\Rightarrow x^{2} + 8x + \frac{8 \times 15}{6} - 5 = 0 $$
$$x^{2} + 8x + 15 = 0 $$
$$x = - 3, - 5 $$
$$\alpha = - 5,\alpha + 2 = - 3 $$
$$\alpha + n = 7 - 5 = 2 $$
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