Question

The value of the expression

$(1+\frac{1}{\omega })(1+\frac{1}{{\omega }^2})+(2+\frac{1}{\omega })(2+\frac{1}{{\omega }^2})+(3+\frac{1}{{\omega }^2})(3+\frac{1}{{\omega }^2})+.............+(n+\frac{1}{\omega })(n+\frac{1}{{\omega }^2})$

($\omega$ is the root of unity) 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)

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

$a_n=(n+\frac{1}{\omega })(n+\frac{1}{{\omega }^2})$

$\sum a_n=\sum (n+\frac{1}{\omega })(n+\frac{1}{{\omega }^2})$

$=\sum \left(\frac{\omega n+1}{\omega }\right)\left(\frac{{\omega }^{2}n+1}{{\omega }^2}\right)$

$=\sum \frac{1}{{\omega }^3}({\omega }^3{n}^2+\omega n+{\omega }^{2}n+1)$

$=\sum n^2+(\omega +{\omega }^2)n+1..............[{\omega }^3=1]$

$=\sum n^2+(-1)n+1................[1+\omega +{\omega }^2=0]$

$=\sum n^2-n+1$

$=\sum n^2-\sum n+\sum 1$

$=n[\frac{(n+1)(2n+1)}{6}-\frac{(n+1)}{2}+1]$

$=n\left[\frac{2n^2+2n+n+1-3n-3+6}{6}\right]$

$=n\left[\frac{2n^2+4}{6}\right]$

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