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If `omega` is a complex nth root of unity, then `sum_(r=1)^n(ar+b)omega^(r-1)` is equal to
A..`(n(n+1)a)/2`
B. `(n b)/(1+n)`
C. `(n a)/(omega-1)`
D. none of these

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To solve the problem, we need to evaluate the summation: \[ S = \sum_{r=1}^{n} (ar + b) \omega^{r-1} \] where \(\omega\) is a complex \(n\)th root of unity. ...
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