Question

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

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

Answer 1

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

$1(2-\omega )(2-{\omega }^2)+2(3-\omega )(3-{\omega }^2)+\cdots \cdots +(n-1)(n-\omega )(n-{\omega }^2)$
$k^{th}$ term of the expression is
$t_k=(k-1)(k-\omega )(k-{\omega }^2)$
$S_n=\sum _{k=2}^n(k-1)(k-\omega )(k-{\omega }^2)$

$=\sum _{k=2}^n(k-1)(k^2+k+1)=\sum _{k=2}^n(k^3-1)$

$=(2^3+3^3+\cdots .+n^3)-(n-1)$

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

Alternate solution
Putting value of $n=2$
$1(2-\omega )(2-{\omega }^2)=4-2\omega -2{\omega }^2+{\omega }^3=7$
Now putting value of $n$ in option and checking which one satisfies
So we get
${\left(\frac{n(n+1)}{2}\right)}^2-n$