Question

The value of the expression
$2(1+\frac{1}{\omega })(1+\frac{1}{{\omega }^2})+3(2+\frac{1}{\omega })(2+\frac{1}{{\omega }^2})+4(3+\frac{1}{\omega })(3+\frac{1}{{\omega }^2})+\cdots +(n+1)(n+\frac{1}{\omega })(n+\frac{1}{{\omega }^2}),$
where $\omega$ is an imaginary cube root of unity, is

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

Answer 1

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

$t_n=(n+1)(n+\frac{1}{\omega })(n+\frac{1}{{\omega }^2})$
$=(n+1)(n+{\omega }^2)(n+\omega )$
$=(n+1)(n^2-n+1)$
$=n^3+1$
$\therefore S_n=\sum _{r=1}^n{t}_r=\sum _{r=1}^n(r^3+1)={\left(\frac{n(n+1)}{2}\right)}^2+n$
$=\frac{n^2(n+1{)}^2}{4}+n$