Question

The value of the expression
$1\cdot (2-\omega )(2-{\omega }^2)+2(3-\omega )(3-{\omega }^2)+....+(n-1)(n-\omega )(n-{\omega }^2)$,where $\omega$ is an imaginary cube root of unity, 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 $r\left[\right(r+1)-\omega ]\left[\right(r+1)-{\omega }^2]$
$=r[{(r+1)}^{}-(\omega +{\omega }^2)(r+1)+{\omega }^2]$
$=r(r^2+3r+3)=r^3+3r^2+3r$
Thus sum of the given series
$={\sum }_{r=1}^{n-1}(r^3+3r^2+3r)=\frac{1}{4}{(n-1)}^2{n}^2+3\frac{1}{6}(n-1)n(2n-1)+3.\frac{1}{2}(n-1)n$
$=\frac{1}{4}(n-1)n(n^2+3n+4)$