AP EAMCET · Maths · Sequences and Series
If \(\omega\) is a complex cube root of unity, then for any \(n>1\), \(\sum_{r=1}^{n-1} r(r+1-\omega)\left(r+1-\omega^2\right)=\)
- A \(\frac{n^2(n+1)^2}{4}\)
- B \(\frac{n(n+1)(2 n+1)}{6}\)
- C \(\frac{n(n-1)}{4}\left(n^2+3 n+4\right)\)
- D \(\frac{n(n+1)(2 n+1)}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{n(n-1)}{4}\left(n^2+3 n+4\right)\)
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