If 1 + x 2 = 3 x , then ∑ n = 1 24 x n - 1 x n 2 is equal to

Given, $x^2-\sqrt{3}x+1=0$

$\Rightarrow x=\frac{\sqrt{3}\pm \sqrt{3-4}}{2}=\frac{\sqrt{3}\pm i}{2}=\cos \frac{\pi }{6}\pm i\sin \frac{\pi }{6}$

$\Rightarrow x^n=\cos \frac{n\pi }{6}\pm i\sin \frac{n\pi }{6}$

And $\frac{1}{x^n}=\cos \frac{n\pi }{6}\pm i\sin \frac{n\pi }{6}$

$\therefore x^n-\frac{1}{x^n}=\pm 2i\sin \frac{n\pi }{6}$

$\Rightarrow {\left(x^n-\frac{1}{x^n}\right)}^2=-4{\sin }^2\frac{n\pi }{6}$            $\left[i^2=-1\right]$

Hence, $\sum _{n=1}^{24}{\left(x^n-\frac{1}{x^n}\right)}^2$

$=-4\left[{\sin }^2\frac{\pi }{6}+{\sin }^2\frac{2\pi }{6}+...+{\sin }^2\frac{24\pi }{6}\right]$

$=-4\left[{\sin }^2{30}^0+{\sin }^2{60}^0+{\sin }^2{90}^0+{\sin }^2{120}^0+{\sin }^2{150}^0+{\sin }^2{180}^0...+{\sin }^2{720}^0\right]$

$=-4\times 4\left[{\sin }^2{30}^0+{\sin }^2{60}^0+{\sin }^2{90}^0+{\sin }^2{120}^0+{\sin }^2{150}^0+{\sin }^2{180}^0\right]$

$=-4\times 4\left[\frac{1}{4}+\frac{3}{4}+1+\frac{3}{4}+\frac{1}{4}+0\right]$

$=-4\times 4\left[3\right]$

$=-4\left(12\right)=-48$