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Question

The value of the expression 1(2ω)(2ω2)+2(3ω)(3ω2)++(n1)(nω)(nω2) is where ω is an imaginary cube root of unity is

A
(n(n+1)2)2
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B
(n(n+1)2)2n
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C
(n(n+1)2)2+n
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D
(n(n+1)2)2+2n
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Solution

The correct option is B (n(n+1)2)2n
1(2ω)(2ω2)+2(3ω)(3ω2)++(n1)(nω)(nω2)
kth term of the expression is
tk=(k1)(kω)(kω2)
Sn=nk=2(k1)(kω)(kω2)

=nk=2(k1)(k2+k+1)=nk=2(k31)

=(23+33+.+n3)(n1)

=(n(n+1)2)2n

Alternate solution
Putting value of n=2
1(2ω)(2ω2)=42ω2ω2+ω3=7
Now putting value of n in option and checking which one satisfies
So we get
(n(n+1)2)2n

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