JEE MAIN - Mathematics (2020 - 8th January Evening Slot - No. 7)
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 :
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 :
x2 + 101x + 100 = 0
x2 + 102x + 101 = 0
x2 – 102x + 101 = 0
x2 – 101x + 100 = 0
Explanation
$$\alpha = {{ - 1 + i\sqrt 3 } \over 2}$$ = $$\omega $$
$$a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}} $$
= $$\left( {1 + \omega } \right)\sum\limits_{k = 0}^{100} {{\omega ^{2k}}} $$
= $$\left( {1 + \omega } \right){{1\left( {1 - {{\left( {{\omega ^2}} \right)}^{101}}} \right)} \over {1 - {\omega ^2}}}$$
= $${{1 - {\omega ^{202}}} \over {1 - \omega }}$$
= $${{1 - \omega } \over {1 - \omega }}$$ = 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
$$a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}} $$
= $$\left( {1 + \omega } \right)\sum\limits_{k = 0}^{100} {{\omega ^{2k}}} $$
= $$\left( {1 + \omega } \right){{1\left( {1 - {{\left( {{\omega ^2}} \right)}^{101}}} \right)} \over {1 - {\omega ^2}}}$$
= $${{1 - {\omega ^{202}}} \over {1 - \omega }}$$
= $${{1 - \omega } \over {1 - \omega }}$$ = 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
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