Question

The value of the expression $1\cdot (2-\omega )(2-{\omega }^2)+2\cdot (3-\omega )(3-{\omega }^2)+\cdots +(n-1)(n-\omega )(n-{\omega }^2)$, 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

Answer: (B)

$\frac{1}{4}(n-1)n(n^2+3n+4)$

$r^{th}$ term of the given series
$=r[(r+1)-\omega ][(r+1)-{\omega }^2]$
$=r[{(r+1)}^2-(\omega +{\omega }^2)(r+1)+{\omega }^3]$
$=r[{(r+1)}^2-(-1)(r+1)+1]$
$=r[r^2+3r+3]=r^3+3r^2+3r$
Thus, sum of the given series
$=\sum _{r=1}^{(n-1)}(r^3+3r^2+3r)$
$=\frac{1}{4}{(n-1)}^2{n}^2+3\cdot \frac{1}{6}(n-1)n(2n-1)+3\cdot \frac{1}{2}(n-1)n$
$=\frac{1}{4}(n-1)n(n^2+3n+4)$