"Let $\alpha = {{ - 1 + i\sqrt 3 } \over 2}$. If $a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}}$ and $b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}}$, then a and b are the roots of the quadratic equation :"

Question

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"

Let $\alpha = {{ - 1 + i\sqrt 3 } \over 2}$.
If $a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}}$ and
$b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}}$, then a and b are the roots of the quadratic equation :

Solution

About this question

Drill 25 more like these. Every day. Free.

Let $\alpha = {{ - 1 + i\sqrt 3 } \over 2}$. If $a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}}$ and $b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}}$, then a and b are the roots of the quadratic equation :

Solution

About this question

More Complex Numbers and Quadratic Equations questions

Drill 25 more like these. Every day. Free.

Let $\alpha = {{ - 1 + i\sqrt 3 } \over 2}$.
If $a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}}$ and
$b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}}$, then a and b are the roots of the quadratic equation :