1. Tardigrade
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  3. Mathematics
  4. Let α, β denote the cube roots of unity other than 1 and α ≠ β. Let displaystyle ∑n=0302(-1)n((α/β))n. Then the value of s is

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

WBJEEWBJEE 2014Complex Numbers and Quadratic Equations

Solution:

Case I Let $\alpha=\omega$ and $\beta=\omega^{2}$
$\therefore S=\displaystyle\sum_{n=0}^{302}(-1)^{n}\left(\frac{\omega}{\omega^{2}}\right)^{n}$
$=\displaystyle\sum_{n=0}^{302}(-1)^{n}\left(\omega^{2}\right)^{n}$
$=1-\omega^{2}+\omega^{4}-\omega^{6} +\omega^{8}-\omega^{10}+\omega^{12} + \ldots+\omega^{600}-\omega^{602}+\omega^{604} $
$=1-\omega^{2}+\omega-1+\omega^{2}-\omega+1+ \ldots +1-\omega^{2}+\omega $
$=0+\ldots+1-\omega^{2}+\omega$
$=-\omega^{2}-\omega^{2}=-2 \omega^{2}$
$\left[\because 1+\omega+\omega^{2}=0\right]$
Case II Let $\alpha=\omega^{2}$ and $\beta=\omega$
$\therefore S=\displaystyle\sum_{n=0}^{302}(-1)^{n}\left(\frac{\omega^{2}}{\omega}\right)^{n}$
$=\displaystyle\sum_{n=0}^{302}(-1)^{n}\left(\frac{\omega^{4}}{\omega^{3}}\right)^{n}$
$=\displaystyle\sum_{n=0}^{302}(-1)^{n}(\omega)$
$=1-\omega+\omega^{2}-\omega^{3}+\omega^{4}-\omega^{5}+\omega^{6}-\ldots+\omega^{300}-\omega^{301}+\omega^{302}$
$=1-\omega+\omega^{2}-1+\omega-\omega^{2}+1 -\ldots+1-\omega+\omega^{2} $
$=0+\ldots+1+\omega^{2}-\omega$
$=-\omega-\omega=-2 \omega$