Q132 / 200 · WBJEE 2014 · Mathematics
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
- Correct answerAeither $-2 \omega$ or $-2 \omega^{2}$
- Beither $-2 \omega$ or $2 \omega^{2}$
- Ceither $2 \omega$ or $-2 \omega^{2}$
- Deither $2 \omega$ or $2 \omega^{2}$
Solution
Case I Let $\alpha=\omega$ and $\beta=\omega^{2}$
$S=\sum_{n=0}^{302}(-1)^{n}\left(\frac{\omega}{\omega^{2}}\right)^{n}$
$=\sum_{n=0}^{302}(-1)^{n}\left(\omega^{2}\right)^{n}$
$=1-\omega^{2}+\omega^{4}-\omega^{6}+\omega^{8}-\omega^{10}+\omega^{12}$
$\quad+\ldots+\omega^{600}-\omega^{602}+\omega^{604}$
$=1-\omega^{2}+\omega-1+\omega^{2}-\omega+1+\ldots$
$=0+\ldots+1-\omega^{2}+\omega$
$=-\omega^{2}-\omega^{2}=-2 \omega^{2} \quad\left[\because 1+\omega+\omega^{2}=0\right]$
Case II Let $\alpha=\omega^{2}$ and $\beta=\omega$
$\therefore S=\sum_{n=0}^{\infty_{2}}(-1)^{n}\left(\frac{\omega^{2}}{\omega}\right)^{n}$
$=\sum_{n=0}^{302}(-1)^{n}\left(\frac{\omega^{4}}{\omega^{3}}\right)^{n}$
$=\sum_{n=0}^{302}(-1)^{n}(\omega)$
$=1-\omega+\omega^{2}-\omega^{3}+\omega^{4}-\omega^{5}+\omega^{6}-\ldots$
$+\omega^{300}-\omega^{301}+\omega^{302}$
$=1-\omega+\omega^{2}-1+\omega-\omega^{2}+1$
$=0+\ldots+1+\omega^{2}-\omega$
$=-\omega-\omega=-2 \omega$
Hence, the correct answer is option: 'A'.