Question

If $\omega$ is a complex cube root of unity, then for any $n>1,\sum _{r=1}^{n-1}r(r-1-\omega )\left(r+1-{\omega }^2\right)=$

• A. $\frac{n^2(n+1{)}^2}{4}$
• B. $\frac{n(n+1)(2n+1)}{6}$
• C. $\frac{n(n-1)}{4}\left(n^2+3n+4\right)$
• D. $\frac{n(n+1)(2n+1)}{4}$

Answer 1

Answer: (C)

Consider,
$\sum _{I=1}^{n-1}r(r+1-w)\left(r+1-w^2\right)$
$=\sum _{I=1}^{n-1}r\left((r+1{)}^2-(r+1)\left(w+w^2\right)+w^3\right)$
$=\sum _{I=1}^{n-1}r\left((r+1{)}^2-(r+1)(-1)+1\right)$
$[\because w^3=1\text{ and }1+w+w^2=01$
$=\sum _{I=1}^{n-1}r\left(r^2+3r+3\right)$
$=\sum _{I=1}^{n-1}\left(r^3+3r^2+3r\right)$
$=\sum _{I=1}^{n-1}{r}^3+3\sum _{I=1}^{n-1}{r}^2+3\sum _{I=1}^{n-1}r$
$={\left(\frac{(n-1)n}{2}\right)}^2+\frac{3(n-1)(n)(2n-1)}{6}+\frac{3(n-1)n}{2}$
$\frac{n^2(n-1{)}^2}{4}+\frac{n(n-1)(2n-1)}{2}+\frac{3}{2}n(n-1)$
$=\frac{n(n-1)}{4}[n(n-1)+2(2n-1)+6]$
$=\frac{n(n-1)}{4}\left[n^2-n+4n-2+6\right]$
$=\frac{n(n-1)}{4}\left[n^2+3n+4\right]$