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If ω is an imaginary cube root of unity, then the value of. Image attached

sohan goswami , 7 Years ago
Grade 12th pass
anser 2 Answers
Amit

Last Activity: 7 Years ago

Hi there. If you want to solve such questions fast replace value of n in ques and well as options .. here replace n=1. To get original statement as zero .. also by replacing n=1 in option A we are getting it zero.. Thats a trick or shortcut..
Samyak Jain

Last Activity: 7 Years ago

Let S = (2 – \omega)(2 – \omega2) + 2(3 – \omega)(3 – \omega2) + …............. +(n – 1)(n – \omega)(n – \omega2)
General term of S, Tr = (r – 1)(r – \omega)(r – \omega2) 
                                   =  (r – 1)[r2 – (\omega + \omega2)r + \omega3]
                                   = (r – 1)(r2 + r + 1)                               [\because 1 + \omega + \omega2 = 0]
                                    = r3 + r2 + r – r2 – r – 1
                                    = r3 – 1
\therefore S = ∑r=2r=n  Tr  =  ∑ r3  –  ∑1  =  [n(n+1)\slash4]2 – n
\therefore S = [{n2 (n\;+\;1)2}\;/\;4]  –  n
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