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Complex Numbers

Section: Mathematics

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Question : 8 of 50

Marks: +1, -0

If

1, \omega, \omega^2

are the cube roots of unity, then

1 \left(2+\;\frac{1}{\omega}\right) \left(2+\;\frac{1}{\omega^2}\right) +2 \left(3+\;\frac{1}{\omega}\right) \left(3+\;\frac{1}{\omega^2}\right) +3 \left(4+\;\frac{1}{\omega}\right) \left(4+\;\frac{1}{\omega^2}\right) +\ldots 10

=

[AP EAPCET 23 May 2025 S1]

$3080$
$3465$
$3175$
$3715$

Solution:

Given,

1, \omega, \omega^2

are cube roots of unity.

Also, we can write the general term (

t_n

) of given expression

t_n\;=(n-1) \left(n+\;\frac{1}{\omega}\right) \left(n+\;\frac{1}{\omega^2}\right)

\;=(n-1) (n+\omega^2) (n+\omega)

\;=(n-1) (n^2+n (\omega+\omega^2) +\omega^3)

\;=(n-1) (n^2-n+1) [ \because 1+\omega+\omega^2=0]

\;= (n^3-2 n^2+2 n-1)

Now,

\sum\limits_{r=1}^{n} r=\;\frac{n(n+1)}{2}

,

\;\sum\limits_{r=1}^{n} r^2=\;\frac{n(n+1)(2n+1)}{6}

\;\sum\limits_{r=1}^{n} r^3= \left(\;\frac{n(n+1)}{2}\right)^2

So, we have;

\;\sum\limits_{n=2}^{11} n= \left(\sum\limits_{n=2}^{11} n-1\right)

\;\;\; =\;\frac{11(12)}{2}-1=66-1=65

\;\sum\limits_{n=2}^{11} n^2- \left(\sum\limits_{n=1}^{11} n^2\right) -1

\;=\;\frac{11(12)(23)}{6}-1=506-1=505

\sum\limits_{n=2}^{11} n^3\;= \left(\sum\limits_{n=1}^{11} n^3\right) -1

\;= \left(\;\frac{11(12)}{2}\right)^2-1=(66)^2-1

\;=4356-=4355

\;\text{ and }\; \sum\limits_{n=2}^{11} 1\;=10

So, final answer

\;=4355-2(505)+2(65)-10

\;=4355-1010+130-10=3465

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