Q93 / 155 · WBJEE 2016 · Mathematics
If $\omega$ is an imaginary cube root of unity, then the value of $(2-\omega)\left(2-\omega^{2}\right)+2(3-\omega)\left(3-\omega^{2}\right)$ $+\ldots+(n-1)(n-\omega)\left(n-\omega^{2}\right)$ is
- Correct answerA$\frac{n^{2}}{4}(n+1)^{2}-n$
- B$\frac{n^{2}}{4}(n+1)^{2}+n$
- C$\frac{n^{2}}{4}(n+1)^{2}$
- D$\frac{n^{2}}{4}(n+1)-n$
Solution
Given,
$(2-\omega)\left(2-\omega^{2}\right)+2(3-\omega)\left(3 \omega^{2}\right)+\ldots+$ $\quad(n-1)(n-\omega)\left(n \omega^{2}\right)$
Now, $\begin{aligned} T_{n} &=(n-1)(n-\omega)\left(n-\omega^{2}\right) \\ &=(n-1)\left(n^{2}-n \omega-n \omega^{2}+\omega^{3}\right) \\ &=(n-1)\left(n^{2}+n+1\right)=n^{3}-1 \\ \therefore \quad S_{n} &=\Sigma T_{n} \\ &=\Sigma n^{3}-\Sigma 1=\frac{n^{2}(n+1)^{2}}{4}-n \end{aligned}$
Hence, the correct answer is option: 'A'.