Question

If $1+x^2=\sqrt{3x}$ then $\sum _{n=1}^{24}{[x^n-\frac{1}{x^n}]}^2$ is equal to

• A. $48$
• B. $-24$
• C. $\pm 48(\omega -{\omega }^2)$
• D. None of these

Answer 1

The correct option is B $-24$

$1-\sqrt{3}x+x^2=0$
$x=\frac{\sqrt{3}\pm i}{2}$
$x=\frac{\sqrt{3}+i}{2}=-i\overline{w}$ and
$\overline{x}=\frac{\sqrt{3}-i}{2}=iw$
Now
$(x^n-\frac{1}{x^n}{)}^2$
$=x^{2n}+\overline{x^{2n}}-2$
$=-\overline{w^{2n}}-w^{2n}-2$
$=-2-[\overline{w^{2n}}+w^{2n}]$
Now
$\overline{w}=w^2$
And
$1+w+w^2=0$
Hence
$-2-[\overline{w^{2n}}+w^{2n}]$
$=-2-[w^{4n}+w^{2n}]$
$=-2-[-1]$
$=-2+1$
$=-1$
Hence
${\sum }_{n=1}^{n=24}(x^n-\frac{1}{x^n}{)}^2$
$=-24$.