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Q.

The value of ∑k=16(sin 2kπ7−icos2kπ7)=

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a

−1

b

0

c

−i

d

i

answer is D.

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Detailed Solution

∑k=16(sin2kπ7−icos2kπ7)=−i∑k=16(cos2kπ7+isin2kπ7) =−i∑k=16ei2πk7=−i[−1+∑k=06ei2kπ7] =−i[−1+0]=i
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The value of ∑k=16(sin 2kπ7−icos2kπ7)=