IIT-JEE 1996Question 1
The value of the expression $1\cdot(2-\omega)(2-\omega^{2})+2\cdot(3-\omega)(3-\omega^{2})+\cdots+(n-1)\cdot(n-\omega)(n-\omega^{2})$, where $\omega$ is an imaginary cube root of unity, is ______.
Complex Numbers
Answer$\dfrac14 n(n-1)\left(n^{2}+3n+4\right)$
Solution
$(m-\omega)(m-\omega^2)=m^2+m+1$, so the sum is $\sum_{k=1}^{n-1}k(k^2+3k+3)=\sum k^3+3\sum k^2+3\sum k=\frac{n^2(n-1)^2}{4}+\frac{n(n-1)(2n-1)}{2}+\frac{3n(n-1)}{2}=\frac14n(n-1)(n^2+3n+4)$.