Question

Let $\alpha$ and $\beta$ be roots of the equation $x^2-x+1=0$. Then the value of $\alpha +\beta +{\alpha }^2+{\beta }^2+\cdots +{\alpha }^{100}+{\beta }^{100}$ is

• A. $-3$
• B. $-1$
• C. $0$
• D. $3$

Answer 1

The correct option is A $-3$

$x^2-x+1=0$
So, $\alpha =-\omega$ and $\beta =-{\omega }^2$, where $\omega$ is the cube root of unity.

Now, $\alpha +\beta +{\alpha }^2+{\beta }^2+\cdots +{\alpha }^{100}+{\beta }^{100}$
$=(\alpha +{\alpha }^2+\cdots +{\alpha }^{100})+(\beta +{\beta }^2+\cdots +{\beta }^{100})$
$=\frac{\alpha (1-{\alpha }^{100})}{1-\alpha }+\frac{\beta (1-{\beta }^{100})}{1-\beta }$
$=\frac{-\omega (1-{\omega }^{100})}{1+\omega }+\frac{-{\omega }^2(1-{\omega }^{200})}{1+{\omega }^2}$
$=\frac{-\omega }{-{\omega }^2}(1-\omega )+\frac{-{\omega }^2}{-\omega }(1-{\omega }^2)$
$=\frac{1}{\omega }(1-\omega )+\omega (1-{\omega }^2)$
$={\omega }^2(1-\omega )+\omega (1-{\omega }^2)$
$={\omega }^2-1+\omega -1$
$=\omega +{\omega }^2-2$
$=-1-2$
$=-3$