Question

$\displaystyle \sum_{k=1}^{6}\left[\sin \frac{2 k \pi}{7}-i \cos \frac{2 k \pi}{7}\right]$ is equal to

• A. -1
• B. 0
• C. #NAME?
• D. i

Answer 1

Answer: (D)

i

Answer 2

Given, $\displaystyle \sum_{k=1}^{6}\left[\sin \frac{2 k \pi}{7}-i \cos \frac{2 \pi k}{7}\right]$
$=\displaystyle \sum_{k=1}^{6}\left[(-i)\left(\cos \frac{2 k \pi}{7}+i \sin \frac{2 \pi k}{7}\right)\right]$
$=(-i) \displaystyle \sum_{k=1}^{6}\left(\cos \frac{2 k \pi}{7}+i \sin \frac{2 \pi k}{7}\right)$
$=(-i) \displaystyle \sum_{k=1}^{6} \alpha^{k}$
Let $\alpha =\cos \frac{2 \pi k}{7}+i \sin \frac{2 \pi k}{7}$
$=(-i)\left(\alpha+\alpha^{2}+\alpha^{3}+\ldots+\alpha^{6}\right)$
It is a GP of which the first term is $\alpha$, number of terms is 6 and the common ratio is $\alpha$.
$\therefore S=(-i) \frac{\alpha\left(1-\alpha^{6}\right)}{1-\alpha}=(-i) \frac{\alpha-\alpha^{7}}{1-\alpha}$
$=(-1) \frac{\alpha-1}{1-\alpha}=i$
$\left[\because \alpha^{7}=\cos 2 \pi+i \sin 2 \pi=1\right]$