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

Verified by Experts

The correct Answer is:

C

Similar Questions

Explore conceptually related problems

The value of sum_(n=0)^(100)i^(n!) equals (where i=sqrt(-1))

Find he value of sum_(r=1)^(4n+7)\ i^r where, i=sqrt(- 1).

if cos (1-i) = a+ib, where a , b in R and i = sqrt(-1) , then

The value of sum_(k=1)^(13)(i^(n)+i^(n+1)) , where i=sqrt(-1) equals :

The value of the sum sum_(n=1) ^(13) (i^(n)+i^(n+1)) , where i = sqrt( - 1) ,equals :

The value of the sume sum_(n=1)^(13) ( i^(n) + i^(n+1)) , where i = sqrt( -1) , equals :

The simplified form of i^(n)+i^(n+1)+i^(n+2)+i^(n+3) is

sum_(i=1)^(n) sum_(j=1)^(i)sum_(k=1)^(j) 1 equals :

(1+i)^(2 n)+(1-i)^(2 n), n in z is