Question

The value of the expression $1(2-\omega )(2-{\omega }^2)+2(3-\omega )(3-{\omega }^2)+......+(n-1).(n-\omega )(n-{\omega }^2)$, where $\omega$ is an imaginary cube root of unit, is:

• A. $\frac{1}{2}(n-1)n(n^2+3n+4)$
• B. $\frac{1}{4}(n-1)n(n^2+3n+4)$
• C. $\frac{1}{2}(n+1)n(n^2+3n+4)$
• D. $\frac{1}{4}(n+1)n(n^2+3n+4)$

Answer 1

The correct option is B $\frac{1}{4}(n-1)n(n^2+3n+4)$

We have

$(Z-1)(Z-W)(Z-W^2)=Z^3-1$

$\therefore |(2-W)(2-W^2)+2(3-W)(3-W^2)|+---+(n-1)(n-W)(n-W)$

$=\sum _{r=2}^n(r-1)(r-W)(r-W^2)$

$=\sum _{r=2}^n{r}^3-\sum _{r=2}^{n}1$

$=\sum _{r=2}^n\left(r^3\right)-1-\left(\sum _{r=2}^{n}1\right)$

$={\left\{\frac{n(n+1)}{2}\right\}}^2-1-(n-1)$

$=\frac{n^2(n+1{)}^2}{4}-n$

$=\frac{n^4+2n^3+n^2-4n}{4}$

$=\frac{(n-1)\cdot n(n^4+3n+4)}{4}$