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

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

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

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

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

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

2549. If sum i^((2n+1)!)=a+ib, where i=sqrt(-1, then a-b, is )

Prove that i^(n)+i^(n+1)+i^(n+2)+i^(n+3)=0 for all n inN