Question

The value of the expression $2(1+\omega )(1+{\omega }^2)+3(2{\omega }^2+1)(2{\omega }^2+1)+4(3\omega +1)(3{\omega }^2+1)+...+(n\omega +1)(n{\omega }^2+1)$ equals

• A. $\frac{n^2{(n+1)}^2}{4}$
• B. $\frac{n^2{(n+1)}^2}{4}+n$
• C. $\frac{n^2{(n+1)}^2}{4}-n$
• D. None of these

Answer 1

The correct option is B $\frac{n^2{(n+1)}^2}{4}+n$

Let $w$ be the cube root of unity.
$\therefore w^3=1\&1+w+w^2=0$
$z=2(1+w)(1+w^2)+3(1+2w)(1+2w^2)+4(1+3w)(1+3w^2).....+(n+1)(1+nw)(1+n{w}^2)$
$\Rightarrow z={\sum }_{r=1}^n(r+1)(1+rw)(1+r{w}^2)={\sum }_{r=1}^n(r+1)[1+r(w+w^2)+r^2{w}^3]$ ...{$\because w^3=1\&1+w+w^2=0$}
$\Rightarrow z={\sum }_{r=1}^n(r+1)(1-r+r^2)$
$\Rightarrow z={\sum }_{r=1}^n(r^3+1)$
$\Rightarrow z=\frac{n^2{(n+1)}^2}{4}+n$
Hence, option 'B' is correct.