Question

Value of $\displaystyle\sum_{ k =1}^{6}\left(\frac{2 k \pi}{7}\right)- i \cos \left(\frac{2 k \pi}{7}\right)$ is equal to.

• A. -1
• B. 1
• C. 0
• D. none of these

Answer 1

Answer: (D)

none of these

Answer 2

$x =\displaystyle\sum_{ k =1}^{6} \sin \left(\frac{2 k \pi}{7}\right)- i \cos \left(\frac{2 k \pi}{7}\right)$ $x =\frac{1}{ i } \displaystyle\sum_{ k =1}^{6} i \sin \left(\frac{2 k \pi}{7}\right)- i ^{2} \cos \left(\frac{2 k \pi}{7}\right)$ $x =\frac{1}{ i }\displaystyle \sum_{ k =1}^{6} \cos \left(\frac{2 k \pi}{7}\right)+ i \sin \left(\frac{2 k \pi}{7}\right)$ $\left[\because i ^{2}=-1\right]$ Now, $\cos \theta+i \sin \theta=e^{i \theta}$ $\therefore x =\frac{1}{ i }\displaystyle \sum_{ k =1}^{6} e ^{ i }\left[\frac{2 k \pi}{7}\right]$ $=\frac{1}{ i } e ^{ i } \frac{2 k \pi}{7}[1+2+3+4+5+6]$ $=\frac{1}{ i } e ^{ i } \frac{2 k \pi}{7} \times 21$ $=\frac{1}{ i } e ^{ i 6 \pi}$ $=\frac{1}{ i }[\cos 6 \pi+ i \sin 6 \pi]$ $=\frac{1}{ i }(1+0)$ $=\frac{1}{ i }=\frac{1}{ i } \times \frac{ i }{ i }$ $=\frac{ i }{ i ^{2}}=- i$