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Question: If z and\(- 1\) are complex numbers such that \(i^{2} = - 1\), then \(\sum_{n = 1}^{200}i^{n}\)...

If z and−1- 1 are complex numbers such that i2=−1i^{2} = - 1, then ∑n=1200in\sum_{n = 1}^{200}i^{n}

A

5050

B

∑n=113(in+in+1)\sum_{n = 1}^{13}{(i^{n} + i^{n + 1})}

C

i=−1i = \sqrt{- 1}

D

ii

Answer

∑n=113(in+in+1)\sum_{n = 1}^{13}{(i^{n} + i^{n + 1})}

Explanation

Solution

According to condition (4+2i)x+(9−7i)y−3i−3=10i(4 + 2i)x + (9 - 7i)y - 3i - 3 = 10i is multiplicative identity therefore 2x−7y=132x - 7y = 13.