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VITEEE 2015 Paper

Section: Mathematics

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Question : 89 of 120

Marks: +1, -0

If

x^2

+ x + 1 = 0, then the value of

\Sigma^{\underset{n=1}{6}} \left(x^n+\frac{1}{x^n}\right)^2

is

13
12
9
14

Solution:

Given equation is

x^2

+ x + 1 = 0

⇒ x = ω and x =

\omega^2

Case I : When x = ω

Then

\Sigma^{\underset{n=1}{6}} \left[x^n+\frac{1}{x^n}\right]^2

=

\Sigma^{\underset{n=1}{6}} [\omega^2+\omega^{2n}]^2

[Since 1/ω =

\omega^2

]

=

(\omega \omega^2)^2 + (\omega^2+\omega^4)^2

+

(\omega^3+\omega^6)^2 + (\omega^4+\omega^8)^2

+

(\omega^5+\omega^{10})^2

+

(\omega^6+\omega^{12})^2

=

9-1)^2 + (-1)^2 + (2)^2 + (-1)^2

+

(-1)^2 + (2)^2

= 12

Case II: When x =

\omega^2

Then

\Sigma^{\underset{n=1}{6}} \left[x^n+\frac{1}{x^n}\right]^2

=

\Sigma^{\underset{n=1}{6}} [\omega^2+\omega^{2n}]^2

[Since 1/

\omega^2

= ω] = 12

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