Question:
If $\omega$ is an imaginary cube root of unity, then the value of $\left(2-\omega\right)\left(2-\omega^{2}\right)+2\left(3-\omega\right)\left(3-\omega^{2}\right)+.....+\left(n-1\right)\left(n-\omega\right)\left(n-\omega^{2}\right)$
If $\omega$ is an imaginary cube root of unity, then the value of $\left(2-\omega\right)\left(2-\omega^{2}\right)+2\left(3-\omega\right)\left(3-\omega^{2}\right)+.....+\left(n-1\right)\left(n-\omega\right)\left(n-\omega^{2}\right)$
Updated On: Aug 9, 2024
- $\frac{n^{2}}{4}\left(n+1\right)^{2}-n$
- $\frac{n^{2}}{4}\left(n+1\right)^{2}+n$
- $\frac{n^{2}}{4}\left(n+1\right)^{2}$
- $\frac{n^{2}}{4}\left(n+1\right)-n$
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The Correct Option is A
Solution and Explanation
$\displaystyle \sum_{r=2}^n\left(r-1\right)\left(r-\omega^{2}\right)=\displaystyle \sum_{r=2}^n(r^3-1)=$$\left[\frac{n^{2}\left(n+1\right)^{2}}{4}-1\right]-\left(n-1\right)=\frac{n^{2}\left(n+1\right)^{2}}{4}-n$
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Concepts Used:
Complex Numbers and Quadratic Equations
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