• # question_answer Let $\omega$ is an imaginary cube roots of unity then the value of$2(\omega +1)({{\omega }^{2}}+1)+3(2\omega +1)(2{{\omega }^{2}}+1)+.....$$+(n+1)(n\omega +1)(n{{\omega }^{2}}+1)$  is [Orissa JEE 2002] A) ${{\left[ \frac{n(n+1)}{2} \right]}^{2}}+n$ B) ${{\left[ \frac{n(n+1)}{2} \right]}^{2}}$ C) ${{\left[ \frac{n(n+1)}{2} \right]}^{2}}-n$ D) None of these

$2\,(\omega +1)\,({{\omega }^{2}}+1)+3(2\omega +1)\,(2{{\omega }^{2}}+1)+......$$+(n+1)\,(n\omega +1)\,\,(n{{\omega }^{2}}+1)$ = $\sum\limits_{r=1}^{n}{(r+1)\,(r\omega +1)\,\,(r{{\omega }^{2}}+1)}$ = $\sum\limits_{r=1}^{n}{(r+1)\,({{r}^{2}}{{\omega }^{3}}+r\omega +r{{\omega }^{2}}+1)}$ = $\sum\limits_{r=1}^{n}{(r+1)\,({{r}^{2}}-r+1)}$= $\sum\limits_{r=1}^{n}{({{r}^{3}}-{{r}^{2}}+r+{{r}^{2}}-r+1)}$ = $\sum\limits_{r=1}^{n}{({{r}^{3}})+\sum\limits_{r=1}^{n}{(1})}$ = ${{\left[ \frac{n(n+1)}{2} \right]}^{2}}+n$.