The value of the expression 1⋅(2−ω)(2−ω2)+2⋅(3−ω)(3−ω2)+....+(n−1)⋅(n−ω)(n−ω2), 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}2−n
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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}2−n Let Sn=1⋅(2−ω)(2−ω2)+2⋅(3−ω)(3−ω2)+....+(n−1)⋅(n−ω)(n−ω2) Tn=(n−1)(n−ω)(n−ω2) =n3−n2(ω2+ω+1)+n(ω3+ω+ω)−ω3 As we know, (ω2+ω+1)=0,ω3=1 Tn=n3−1 Sn=n∑n=2(n3−1) =n∑n=1n3−n∑n=11−(13−1) ={n(n+1)2}2−n