Question

If $x^2 - x + 1 = 0,$ then the value of $ \displaystyle\sum^{5}_{n=1} \left(x^{n} + \frac{1}{x^{n}}\right)^{2}$

• A. 10
• B. 8
• C. 6
• D. 4

Answer 1

Answer: (B)

We have, $x^2 - x + 1 = 0$
$x = \frac{1\pm\sqrt{1-4}}{2} = \frac{1}{2} \pm \frac{\sqrt{3} i}{2} $
i.e., $x = \omega, \omega^{2}$
$\therefore \:\:\: \displaystyle\sum^{5}_{n=1} \left(x^{n} + \frac{1}{x^{n}}\right)^{2} = \displaystyle\sum ^{5}_{n=1} \left(\omega^{n} + \frac{1}{\omega^{n}}\right)^{2}$
$= \displaystyle\sum_{n=1}^{5} \left[\omega^{2n} + \frac{1}{\omega ^{2n} } + 2\right]$
$ = 5\left(2\right) + \omega ^{2} + \frac{1}{\omega ^{2}} +\omega ^{4} + \frac{1}{\omega ^{4}}$
$ + \omega ^{6} + \frac{1}{\omega ^{6}} + \omega ^{8} + \frac{1}{\omega ^{8}} + \omega ^{10} + \frac{1}{\omega ^{10}} $
$ = 10+ \frac{\omega ^{4} + 1}{\omega ^{2}} +\frac{\omega ^{8} + 1}{\omega ^{4}} + \frac{\omega ^{12} + 1}{\omega ^{6}}$
$ + \frac{\omega ^{16} + 1}{\omega ^{8}} + \frac{\omega ^{20} + 1}{\omega ^{10}} $
$ = 10 + \frac{\omega +1}{\omega ^{2}} + \frac{\omega^{2} + 1}{\omega } + \frac{1+1}{1} +\frac{\omega +1}{\omega ^{2}} + \frac{\omega^{2} + 1}{ \omega } $
$ = 10 -\frac{\omega ^{2}}{\omega ^{2}} = \frac{\omega }{\omega } + 2 - \frac{\omega ^{2}}{\omega ^{2}} - \frac{\omega }{\omega } $
$ = 10 - 1 - 1 + 2 - 1 -1 =8$.