# If ${1+x^2=\sqrt3x}$, then $\sum_{n=1}^{24}(x^n-\frac{1}{x^n})$

If ${1+x^2=\sqrt3x}$, then

$\sum_{n=1}^{24}(x^n-\frac{1}{x^n}) \hspace{5mm} ?$

The attempt at solving:

First on solving the quadratic, $$x = \frac{\sqrt3+i}{2} \hspace{5mm} \mbox{ or } \hspace{5mm} x=\frac{\sqrt3-i}{2}$$ Then I put them in Euler's form: $${e^{i\frac{\pi}6}} \hspace{5mm} \& \hspace{5mm} {e^{i\frac{-\pi}6}}.$$ Then the summation is then: $$\sum_{n=1}^{24}({e^{i\frac{npi}6}}-{e^{i\frac{-npi}6})} = \sum_{n=1}^{24}[2i \,{\sin(n\theta)]}$$

EDIT: Earlier I thoght my answer was wrong as it didn't match with the textbook but I have got it now.

• Comments are not for extended discussion; this conversation has been moved to chat. – Jyrki Lahtonen Dec 6 '16 at 8:24

You correctly solved that $x=e^{\pm \pi i/6}$. This implies that $x^{24}=e^{4\pi i}=1$.
By the formula for a geometric sum $$\sum_{n=1}^{24}x^n=\frac{x-x^{25}}{1-x}=\frac{x(1-x^{24})}{1-x}=\frac{x(1-1)}{1-x}=0.$$
I leave it to you to calculate the sum $-\sum_{n=1}^{24}x^{-n}$. It seems to me that your textbook has the wrong answer.