Home
Class 12
MATHS
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

2

B

3

C

5

D

6

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Similar Questions

Explore conceptually related problems

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=4)^(100)i^(r!)+prod_(r=1)^(101)i^(r)=a+ib, " where " i=sqrt(-1) , then a+75b, is

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_(n=1)^(25)i^(n!)=a+ib, " where " i=sqrt(-1) , then a-b, is

The value of sum_(n=0)^(50)i^(2n+n)! (where i=sqrt(-1)) is

The value of sum_(r=-3)^(1003)i^(r)(where i=sqrt(-1)) is

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

If z=(sqrt(3)-i)/2 , where i=sqrt(-1) , then (i^(101)+z^(101))^(103) equals to

sum_(r=0)^(n).^(n)C_(r)4^(r)=..........

The real part of (1-i)^(-i), where i=sqrt(-1) is

sum_(r=1)^n(2r+1)=...... .

If sum_(i=1)^(2n)cos^(-1)x_i=0 then find the value of sum_(i=1)^(2n)x_i