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TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

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Question : 10 of 160

Marks: +1, -0

If

\omega

is a complex cube root of unity, then

\sum\limits_{x=1}^{10}((\omega x+2)(\omega^{2} x+2)-3)=

285
945
1025
705

Solution:

As,

\omega

is complex root of unity, then

1+\omega+\omega^{2}=0, \omega^{3}=1

\therefore

(\omega x+2)(\omega^{2} x+2)-3

=\omega^{3} x^{2}+2\omega x+2\omega^{2} x+4-3

=x^{2}+2x(\omega+\omega^{2})+1 (\because \omega^{3}=1)

= \;\; x^{2}+2x(-1)+1=x^{2}-2x+1

\therefore \;\; \sum\limits_{x=1}^{10}((\omega x+2)(\omega^{2}x+2)-3)

= \;\; \sum\limits_{x=1}^{10}(x^{2}-2x+1)=\sum\limits_{x=1}^{10} x^{2}-2 \sum\limits_{x=1}^{10} x+\sum\limits_{x=1}^{10} 1

\;\; =\;\frac{10(10+1)(2 \times 10+1)}{6}-\;\frac{2 \times 10(10+1)}{2}+10

= \;\; \;\frac{10 \times 11 \times 21}{6}-10 \times 11+10=385-100=285

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