If 1+x^2=sqrt{3} x,then sum_{n=1}^{24}left(x^ | vedclass

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(D) Given,$x^2-\sqrt{3} x+1=0$. Dividing by $x$,we get $x+\frac{1}{x}=\sqrt{3}$. We know that $\left(x^n-\frac{1}{x^n}\right)^2 = \left(x^n+\frac{1}{x

Vedclass · Accepted Answer

$-48$

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(D) Given,$x^2-\sqrt{3} x+1=0$. Dividing by $x$,we get $x+\frac{1}{x}=\sqrt{3}$. We know that $\left(x^n-\frac{1}{x^n}ight)^2 = \left(x^n+\frac{1}{x^n}ight)^2 - 4$. Let $x = e^{i	heta} = \cos 	heta + i \sin 	heta$. Then $x+\frac{1}{x} = 2 \cos 	heta = \sqrt{3}$,so $\cos 	heta = \frac{\sqrt{3}}{2}$,which implies $	heta = \frac{\pi}{6}$. Thus,$x^n = \cos \frac{n\pi}{6} + i \sin \frac{n\pi}{6}$ and $\frac{1}{x^n} = \cos \frac{n\pi}{6} - i \sin \frac{n\pi}{6}$. Then $x^n - \frac{1}{x^n} = 2i \sin \frac{n\pi}{6}$. Therefore,$\left(x^n - \frac{1}{x^n}ight)^2 = (2i \sin \frac{n\pi}{6})^2 = -4 \sin^2 \frac{n\pi}{6}$. Now,$\sum_{n=1}^{24} -4 \sin^2 \frac{n\pi}{6} = -4 \sum_{n=1}^{24} \sin^2 \frac{n\pi}{6}$. Since $\sin^2 \frac{n\pi}{6}$ has a period of $6$,the sum over $24$ terms is $4 	imes \sum_{n=1}^{6} \sin^2 \frac{n\pi}{6}$. $\sum_{n=1}^{6} \sin^2 \frac{n\pi}{6} = \sin^2 \frac{\pi}{6} + \sin^2 \frac{2\pi}{6} + \sin^2 \frac{3\pi}{6} + \sin^2 \frac{4\pi}{6} + \sin^2 \frac{5\pi}{6} + \sin^2 \frac{6\pi}{6} = (\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2 + (1)^2 + (\frac{\sqrt{3}}{2})^2 + (\frac{1}{2})^2 + 0^2 = \frac{1}{4} + \frac{3}{4} + 1 + \frac{3}{4} + \frac{1}{4} + 0 = 3$. Thus,the total sum is $-4 	imes (4 	imes 3) = -48$.

If $1+x^2=\sqrt{3} x$,then $\sum_{n=1}^{24}\left(x^n-\frac{1}{x^n}\right)^2$ is equal to

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