Question

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

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

Answer 1

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

Let $\omega$ be the imaginary cube root of $1$.

We have to find the value of $1(2-\omega )(2-{\omega }^2)+2(3-\omega )(3-{\omega }^2)+....+(n-1)(n-\omega )(n-{\omega }^2)$.

$r^{th}term=\left(r\right)(r+1-\omega )(r+1-{\omega }^2)$

$=(r+1-1)(r+1-\omega )(r+1-{\omega }^2)$

$=(r+1{)}^3-1\cdot \omega \cdot {\omega }^2$

$=(r+1{)}^3-1$

$={\sum }_{r=0}^{n-1}(r+1{)}^3-1$

$={\sum }_{r=0}^n{r}^3-n$

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

Thus the value of $1(2-\omega )(2-{\omega }^2)+2(3-\omega )(3-{\omega }^2)+....+(n-1)(n-\omega )(n-{\omega }^2)$ is $\frac{1}{4}{n}^2(n+1{)}^2-n$.