Question

Let \(\alpha=\dfrac{-1+i\sqrt3}{2}\). If \(a=(1+\alpha)\displaystyle\sum_{k=0}^{100}\alpha^{2k}\) and \(b=\displaystyle\sum_{k=0}^{100}\alpha^{3k}\), then \(a\) and \(b\) are the roots of the quadratic equation :

• A. \(x^2+101x+100=0\)
• B. \(x^2+102x+101=0\)
• C. \(x^2-102x+101=0\)
• D. \(x^2-101x+100=0\)

Hint:

\(\alpha\) is a cube root of unity \(\omega\); use \(1+\omega+\omega^2=0\) and \(\omega^3=1\) to sum the two series.

Answer 1

The Correct Option is C

\(\alpha=\omega\), a non-real cube root of unity: \(\omega^3=1\), \(1+\omega+\omega^2=0\).
\(b=\sum_{k=0}^{100}\omega^{3k}=\sum_{k=0}^{100}1=101\).
For \(a\): the series \(\sum_{k=0}^{100}\omega^{2k}\) has 101 terms. Terms repeat in cycles of 3 with sum \(1+\omega^2+\omega^4=0\); \(101=33\cdot3+2\), so only the last two terms (\(k=99,100\)) remain: \(\omega^{198}+\omega^{200}=1+\omega^2\).
So \(a=(1+\omega)(1+\omega^2)=1+\omega^2+\omega+\omega^3=0+1=1\).
Sum of roots \(=1+101=102\), product \(=101\).
Equation: \(x^2-102x+101=0\).
Answer: (3) \(x^2-102x+101=0\)