Question

If $\omega$ is an imaginary cube root of unity, then
$(1+\frac{1}{\omega })(1+\frac{1}{{\omega }^2})+(1+\frac{2}{\omega })(1+\frac{2}{{\omega }^2})+(1+\frac{3}{\omega })(1+\frac{3}{{\omega }^2})+.....+(1+\frac{n}{\omega })(1+\frac{n}{{\omega }^2})$ is

• A. $\frac{n(n^2+2)}{3}$
• B. $\frac{n(n^2-2)}{3}$
• C. $\frac{n(n^2+1)}{3}$
• D. None of these

Answer 1

Answer: (A)

$\frac{n(n^2+2)}{3}$

$\omega$ is the cube root of unity

$\therefore {\omega }^3=1$

$\frac{1}{\omega }=\frac{{\omega }^3}{\omega }={\omega }^2;\frac{1}{{\omega }^2}=\frac{{\omega }^3}{{\omega }^2}=\omega$
$T_1=(1+{\omega }^2)(1+\omega )$
$=1+\omega +{\omega }^2+{\omega }^3=1(\because 1+\omega +{\omega }^2=0)$
$T_2=(2+{\omega }^2)(2+\omega )$
$=4+2\omega +2{\omega }^2+{\omega }^3=3$
$T_3=(3+{\omega }^2)(3+\omega )$
$=9+3\omega +3{\omega }^2+{\omega }^3=7$
Similarly
$T_4=13$
.
.
.
$T_n=n^2-n+1$
$S_n=\sum t_n=\sum n^2-\sum n+\sum 1$
$=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+n$
$=\frac{n}{6}\{2n^2+3n+1-3n-3+6\}$
$=\frac{n}{6}\{2n^2+4\}=\frac{n}{3}\{n^2+2\}$