If omega is a complex cube root of unity,then | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given that $\omega$ is a complex cube root of unity,we have $1+\omega+\omega^2 = 0$ and $\omega^3 = 1$. Consider the expression $r(r+1-\omega)(r+1

Vedclass · Accepted Answer

$3015$

Vedclass · Answer

(D) Given that $\omega$ is a complex cube root of unity,we have $1+\omega+\omega^2 = 0$ and $\omega^3 = 1$. Consider the expression $r(r+1-\omega)(r+1-\omega^2)$. Expanding this,we get $r[(r+1)^2 - (r+1)(\omega+\omega^2) + \omega^3]$. Since $\omega+\omega^2 = -1$ and $\omega^3 = 1$,the expression becomes $r[(r+1)^2 + (r+1) + 1] = r(r^2+2r+1+r+1+1) = r(r^2+3r+3) = r^3+3r^2+3r$. Now,we calculate the sum $\sum_{r=1}^9 (r^3+3r^2+3r) = \sum_{r=1}^9 r^3 + 3\sum_{r=1}^9 r^2 + 3\sum_{r=1}^9 r$. Using the standard summation formulas: $\sum_{r=1}^9 r^3 = [\frac{9(10)}{2}]^2 = 45^2 = 2025$. $3\sum_{r=1}^9 r^2 = 3 	imes \frac{9(10)(19)}{6} = 3 	imes 285 = 855$. $3\sum_{r=1}^9 r = 3 	imes \frac{9(10)}{2} = 3 	imes 45 = 135$. Adding these values: $2025 + 855 + 135 = 3015$. Thus,the correct option is $D$.

If $\omega$ is a complex cube root of unity,then $\sum_{r=1}^9 r(r+1-\omega)(r+1-\omega^2) = $

Similar Questions

The cube roots of unity when represented on the Argand plane form the vertices of an

If $x+iy=(-1+i\sqrt{3})^{2010}$,then $x$ is

Let $\alpha$ be a root of the equation $1+x^{2}+x^{4}=0$. Then the value of $\alpha^{1011}+\alpha^{2022}-\alpha^{3033}$ is equal to

If $x_n = \cos \frac{\pi}{2^n} + i \sin \frac{\pi}{2^n}$,then $\prod_{n=1}^{\infty} x_n$ is equal to

If $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}\right)^4+\left(\frac{\sqrt{3}-i}{\sqrt{3}+i}\right)^4=r \operatorname{cis} \theta$,then one of the values of $\sqrt{r \operatorname{cis} \theta}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module