1. Tardigrade
  2. Question
  3. Mathematics
  4. If ω is a complex cube root of unity, then for any n>1, displaystyle ∑r=1n-1 r(r-1-ω)(r+1-ω2)=

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

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

AP EAMCETAP EAMCET 2017

Solution:

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