The value of -k-16sin2 k -7-i cos2 k -7 is

The correct option is B i∑ ^6 _k=1 (sin2kπ/7 i cos2kπ/7 )= i ∑ ^6 _k=1 (cos2kπ/7+i sin2kπ/7) = i{∑ ^6 _k=1e^i2kπ/7} = i{e^i2kπ/7+e^i4kπ/7+e^i6kπ/7+e^i8kπ/7+e ...

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Byju's Answer

Standard XI

Mathematics

Euler's Representation

The value of ...

Question

The value of $\sum_{k = 1}^{6} (s i n \frac{2 k π}{7} - i c o s \frac{2 k π}{7})$ is

-1

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-i

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Solution

The correct option is C i
$\sum_{k = 1}^{6} (s i n \frac{2 k π}{7} - i c o s \frac{2 k π}{7}) = - i \sum_{k = 1}^{6} (c o s \frac{2 k π}{7} + i s i n \frac{2 k π}{7})$
$= - i {\sum_{k = 1}^{6} e^{\frac{i 2 k π}{7}}} = - i {e^{\frac{i 2 k π}{7}} + e^{\frac{i 4 k π}{7}} + e^{\frac{i 6 k π}{7}} + e^{\frac{i 8 k π}{7}} + e^{\frac{i 10 k π}{7}} + e^{\frac{i 12 k π}{7}}}$
$= - i ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ e^{\frac{i 2 k π}{7}} \frac{(1 - e^{\frac{i 12 π}{7}})}{1 - e^{\frac{i 2 π}{7}}} ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭$
$= - i {\frac{e^{\frac{i 2 π}{7}} - e^{\frac{i 14 π}{7}}}{1 - e^{\frac{i 2 π}{7}}}} [∵ e^{\frac{i 14 π}{7}} = 1] = - i {\frac{e^{\frac{i 2 π}{7}} - 1}{1 - e^{\frac{i 2 π}{7}}}} = i$

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The value of ∑6k=1(sin2kπ7−i cos2kπ7) is

The value of $\sum_{k = 1}^{6} (s i n \frac{2 k π}{7} - i c o s \frac{2 k π}{7})$ is