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Question

The value of the expression 1(2ω)(2ω2)+2(3ω)(3ω2)+....+(n1)(nω)(nω2), 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
None of the above
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Solution

The correct option is B {n(n+1)2}2n
Let Sn=1(2ω)(2ω2)+2(3ω)(3ω2)+....+(n1)(nω)(nω2)
Tn=(n1)(nω)(nω2)
=n3n2(ω2+ω+1)+n(ω3+ω+ω)ω3
As we know, (ω2+ω+1)=0,ω3=1
Tn=n31
Sn=nn=2(n31)
=nn=1n3nn=11(131)
={n(n+1)2}2n

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