Let A_r = left(x+frac{1}{x}right)^3 cdot left | vedclass

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

Question

(D) Given $x^2+x+1=0$,the roots are $\omega$ and $\omega^2$,where $\omega^3=1$ and $1+\omega+\omega^2=0$. Note that $\omega + \frac{1}{\omega} = \omeg

Vedclass · Accepted Answer

$\frac{1}{7}$

Vedclass · Answer

(D) Given $x^2+x+1=0$,the roots are $\omega$ and $\omega^2$,where $\omega^3=1$ and $1+\omega+\omega^2=0$. Note that $\omega + \frac{1}{\omega} = \omega + \omega^2 = -1$. Thus,the terms in the product $A_r$ are: For $k=1$,$(x+\frac{1}{x})^3 = (-1)^3 = -1$. For $k=2$,$(x^2+\frac{1}{x^2})^3 = (\omega^2+\omega)^3 = (-1)^3 = -1$. For $k=3$,$(x^3+\frac{1}{x^3})^3 = (1+1)^3 = 8$. So,$A_3 = (-1)(-1)(8) = 8$. For $k=4$,$(x^4+\frac{1}{x^4})^3 = (\omega+\omega^2)^3 = -1$. For $k=5$,$(x^5+\frac{1}{x^5})^3 = (\omega^2+\omega)^3 = -1$. For $k=6$,$(x^6+\frac{1}{x^6})^3 = (1+1)^3 = 8$. Thus,$A_6 = A_3 \cdot (-1) \cdot (-1) \cdot 8 = 8^2 = 64$. In general,$A_{3n} = 8^n$. The series is $\sum_{n=1}^{\infty} \frac{1}{A_{3n}} = \sum_{n=1}^{\infty} \frac{1}{8^n} = \frac{1/8}{1-1/8} = \frac{1/8}{7/8} = \frac{1}{7}$.

Let $A_r = \left(x+\frac{1}{x}\right)^3 \cdot \left(x^2+\frac{1}{x^2}\right)^3 \cdot \left(x^3+\frac{1}{x^3}\right)^3 \cdots \left(x^r+\frac{1}{x^r}\right)^3$. If $x^2+x+1=0$,then $\frac{1}{A_3}+\frac{1}{A_6}+\frac{1}{A_9}+\frac{1}{A_{12}}+\cdots \infty =$

Similar Questions

If $\omega$ is a cube root of unity,then the value of $(1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5$ is:

If $x = \cos \theta + i\sin \theta$ and $y = \cos \phi + i\sin \phi$,then ${x^m}{y^n} + {x^{-m}}{y^{-n}}$ is equal to

$\frac{(\cos \alpha + i\sin \alpha )^4}{(\sin \beta + i\cos \beta )^5} = $

Common roots of the equations $z^3 + 2z^2 + 2z + 1 = 0$ and $z^{1985} + z^{100} + 1 = 0$ are

Let $\alpha, \beta$ denote the cube roots of unity other than $1$ and $\alpha \neq \beta$. Let $S = \sum_{n=0}^{\infty} (-1)^{n} \left(\frac{\alpha}{\beta}\right)^{n}$. Then the value of $S$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module