Question

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

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

Answer 1

Answer: (B)

$\frac{1}{4}n[n-1][n^2+3n+4]$

We have:
$1.(4-2w-2w^2+w^3)+2.(9-3w-3w^2+w^3)+3.(16-4w-4w^2+w^3)...till(n-1)terms=1.(4+2+1)+2.(9+3+1)+3.(16+4+1)...till(n-1)terms=1\ast 7+2\ast 13+3\ast 21+...=>t_r=r\left[\right(r+1\left)\right(r+2)+1]forr=1to(n-1)=>S=\sum r^3+3.\sum r^2+3\sum rforr=1to(n-1)={\left(\frac{(n-1)n}{2}\right)}^2+3.\left(\frac{(n-1)\left(n\right)(2n-1)}{6}\right)+3.\left(\frac{(n-1).n}{2}\right)=\frac{1}{4}.n.(n-1).(n^2+3n+4)$
Hence, (B) is correct.