Let =-1+i√32.

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:

Accepted Answer

$$\alpha  = {{ - 1 + i\sqrt 3 } \over 2}$$ = $$\omega $$

$$a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}} $$

= $$\left( {1 + \omega } \right)\sum\limits_{k = 0}^{100} {{\omega ^{2k}}} $$

= $$\left( {1 + \omega } \right){{1\left( {1 - {{\left( {{\omega ^2}} \right)}^{101}}} \right)} \over {1 - {\omega ^2}}}$$

= $${{1 - {\omega ^{202}}} \over {1 - \omega }}$$

= $${{1 - \omega } \over {1 - \omega }}$$ = 1

$$b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}} $$

= $$\sum\limits_{k = 0}^{100} {{\omega ^{3k}}} $$

= 101 [as $${\omega ^3}$$ = 1]

$$ \therefore $$ roots are 101 and 1

Then  equation is = x2
 -- 102x + 101 = 0

Solutions · 1