Question

Let $\alpha ,\beta$ denote the cube roots of unity other than $1$ and $\alpha \ne \beta$. Let $S=\sum _{n=0}^{302}{(-1)}^n{\left(\frac{\alpha }{\beta }\right)}^n$. Then, the value of $S$ is

• A. Either $-2\omega$ or $-2{\omega }^2$
• B. Either $-2\omega$ or $2{\omega }^2$
• C. Either $2\omega$ or $-2{\omega }^2$
• D. Either $2\omega$ or $2{\omega }^2$

Answer 1

Answer: (A)

Either $-2\omega$ or $-2{\omega }^2$

Case I: Let $\alpha =\omega$ and $\beta ={\omega }^2$
$\therefore S=\sum _{n=0}^{302}{(-1)}^n{\left(\frac{\omega }{{\omega }^2}\right)}^n$
$=\sum _{n=0}^{302}{(-1)}^n{\left({\omega }^2\right)}^n$
$=1-{\omega }^2+{\omega }^4-{\omega }^6+{\omega }^8-{\omega }^{10}+{\omega }^{12}+\cdots +{\omega }^{600}-{\omega }^{602}+{\omega }^{604}$
$=1-{\omega }^2+\omega -1+{\omega }^2-\omega +1+\cdots +1-{\omega }^2+\omega$
$=0+\cdots +1-{\omega }^2+\omega$
$=-{\omega }^2-{\omega }^2=-2{\omega }^2$ $[\because 1+\omega +{\omega }^2=0]$

Case II: Let $\alpha ={\omega }^2$ and $\beta =\omega$
$\therefore S=\sum _{n=0}^{302}{(-1)}^n{\left(\frac{{\omega }^2}{\omega }\right)}^n$
$=\sum _{n=0}^{302}{(-1)}^n{\left(\frac{{\omega }^4}{{\omega }^3}\right)}^n$
$=\sum _{n=0}^{302}{(-1)}^n\left(\omega \right)$
$=1-\omega +{\omega }^2-{\omega }^3+{\omega }^4-{\omega }^5+{\omega }^6-\cdots +{\omega }^{300}-{\omega }^{301}+{\omega }^{302}$
$=1-\omega +{\omega }^2-1+\omega -{\omega }^2+1-\cdots +1-\omega +{\omega }^2$
$=0+\cdots +1+{\omega }^2-\omega$
$=-\omega -\omega =-2\omega$