Question

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

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

Answer 1

Answer: (C)

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

$2(1+\omega )(1+{\omega }^{{}^2})+3(2\omega +1)(2{\omega }^2+1)+4(3\omega +1)(3{\omega }^2+1)+\cdots +(n+1)(n\omega +1)(n{\omega }^2+1)$

We know that,

$(a+b)(a+b\omega )(a+b{\omega }^2)=(a+b)(a^2-ab+b^2)$

$=a^3+b^3$

So expression becomes,

$(1^3+1^3)+(2^3+1^3)+\cdots +(n^3+1^3)$

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

Hence, option C.