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Sum of four consecutive powers of i(iota...

Sum of four consecutive powers of i(iota) is zero.
i.e.,`i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0,forall n in I.`
If `sum_(r=-2)^(95)i^(r)+sum_(r=0)^(50)i^(r!)=a+ib, " where " i=sqrt(-1)`, the unit digit of `a^(2011)+b^(2012)`, is

A

(a)2

B

(b)3

C

(c)5

D

(d)6

Text Solution

Verified by Experts

The correct Answer is:
C
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