Question

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

Answer 1

Forming series

We know cube roots of unity are $1,\omega ,{\omega }^2$ and also ${\omega }^3=1$

Now taking $\beta =\omega \&\alpha ={\omega }^2$

$$\begin{gathered} \because S=\sum _{n=0}^{302}{(-1)}^n{\left(\frac{\alpha }{\beta }\right)}^n \\ =\sum _{n=0}^{302}{(-1)}^n{\left(\frac{{\omega }^2}{\omega }\right)}^n \\ =\sum _{n=0}^{302}{(-1)}^n{\left(\omega \right)}^n \\ =1-\omega +{\omega }^2-{\omega }^3+{\omega }^4-{\omega }^5+{\omega }^6-{\omega }^7+{\omega }^8+\dots +{\omega }^{302} \end{gathered}$$

This is a geometrical progression with first term $a=1$ and common ratio $r=-\omega$

Finding sum of GP which is S

Since sum of nth term GP

$$\begin{gathered} \Rightarrow \frac{a(1-r^n)}{(1-r)} \\ \therefore S=\frac{1(1-{(-\omega )}^{303})}{(1+\omega )} \\ =\frac{\left(1{+\left(\omega \right)}^{303}\right)}{(1+\omega )} \\ =\frac{\left(1{+\left({\omega }^3\right)}^{101}\right)}{(1+\omega )} \\ =\frac{(1+1)}{(1+\omega )}\left[\because {\omega }^3=1\right] \\ =\frac{2}{(1+\omega )}\left[\because 1+\omega +{\omega }^2=0\Rightarrow 1+\omega =-{\omega }^2\right] \\ =-\frac{2\times 1}{{\omega }^2} \\ =-\frac{2\times {\omega }^3}{{\omega }^2}\mathbf{}\left[\because {\omega }^3=1\right] \\ =-2\omega \end{gathered}$$

Thus, the required value is $-2\omega$.