Question

If $\omega$ is an imaginary cube root of unity, then the value of $(2-\omega )(2-{\omega }^2)+2(3-\omega )(3-{\omega }^2)+...+(n-1)(n-\omega )(n-{\omega }^2)$ is

• A. $\frac{n^2}{4}(n+1{)}^2-n$
• B. $\frac{n^2}{4}(n+1{)}^2+n$
• C. $\frac{n^2}{4}(n+1{)}^2$
• D. $\frac{n^2}{4}(n+1)-n$

Answer 1

The correct option is A $\frac{n^2}{4}(n+1{)}^2-n$

$n^{th}$ term of the series $(2-\omega )(2-{\omega }^2)+2(3-\omega )(3-{\omega }^2)+...+(n-1)(n-\omega )(n-{\omega }^2)$ is
$T_n=(n-1)(n-\omega )(n-{\omega }^2)$
$=(n-1)(n^2-(\omega +{\omega }^2)n+{\omega }^3)$
$=(n-1)(n^2+n+1)\because 1+{\omega }^2+{\omega }^3=0\text{and}{\omega }^3=1$
$T_n=n^3-1$
Now claculate the summation of $T_n$
$\therefore \sum T_n=\sum n^3-\sum 1$
$={\left(\frac{n(n+1)}{2}\right)}^2-n$
$=\frac{n^2}{4}(n+1{)}^2-n$