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Question: \(i^{2} = - 1,\) then the value of \(\sum_{n = 1}^{200}i^{n}\)is...

i2=−1,i^{2} = - 1, then the value of ∑n=1200in\sum_{n = 1}^{200}i^{n}is

A

50

B

– 50

C

0

D

100

Answer

0

Explanation

Solution

Sol. ∑n=1200in=i+i2+i3+.....+i200=i(1−i200)1−i\sum_{n = 1}^{200}{i^{n} = i + i^{2} + i^{3} + ..... + i^{200} = \frac{i(1 - i^{200})}{1 - i}} (since G.P.)

=i(1−1)1−i=0.= \frac{i(1 - 1)}{1 - i} = 0.