JEE Main 2020 (Online) 8th January Evening ShiftQuestion 1
Let $\alpha=\frac{-1+i\sqrt3}{2}$. If $a=(1+\alpha)\sum_{k=0}^{100}\alpha^{2k}$ and $b=\sum_{k=0}^{100}\alpha^{3k}$, then $a$ and $b$ are the roots of the quadratic equation :
Complex Numbers
Solution
$\alpha=\omega$. $\sum_{k=0}^{100}\omega^{2k}$: 101 terms, groups of 3 sum to 0, $101=33\cdot3+2$, so sum $=1+\omega^2$. $a=(1+\omega)(1+\omega^2)=1+\omega+\omega^2+\omega^3=1$. $b=\sum_{k=0}^{100}1=101$. Roots $1,101$: $x^2-102x+101=0$. Answer: (D).