1.Let α=1+i32\alpha = \dfrac{1 + i\sqrt{3}}{2}α=21+i3. If a=(1+α)∑k=0100α2ka = (1 + \alpha)\sum\limits_{k=0}^{100} \alpha^{2k}a=(1+α)k=0∑100α2k and b=∑k=0100α3kb = \sum\limits_{k=0}^{100} \alpha^{3k}b=k=0∑100α3k, then aaa and bbb are the roots of the quadratic equation:a.x2−101x+100=0x^2 - 101x + 100 = 0x2−101x+100=0b.x2−102x+101=0x^2 - 102x + 101 = 0x2−102x+101=0c.x2+101x−1000=0x^2 + 101x - 1000 = 0x2+101x−1000=0d.x2+102x+101=0x^2 + 102x + 101 = 0x2+102x+101=0Login to continueOnly logged in users canattempt or see the solution.
