simplify : `i^(n+100)+i^(n+50)+i^(n+48)+i^(n+46)`

Simplify the following : (i) 1+ i^(5)+i^(10)+i^(15) (ii) (1+i)^(4)+(1+(1)/(i))^(4) (iii) i^(n)+i^(n+1)+i^(n+2)+i^(n+3)

For positive integer n_1,n_2 the value of the expression (1+i)^(n1) +(1+i^3)^(n1) (1+i^5)^(n2) (1+i^7)^(n_20), where i=sqrt-1, is a real number if and only if (a) n_1=n_2+1 (b) n_1=n_2-1 (c) n_1=n_2 (d) n_1 > 0, n_2 > 0

Find the value of i^n+i^(n+1)+i^(n+2)+i^(n+3) for all n in Ndot

Find the value of i^n+i^(n+1)+i^(n+2)+i^(n+3) for all n in Ndot

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

Simplify the following: (i)\ (3^n\ xx\ 9^(n+1))/(3^(n-1)\ xx\ 9^(n-1)) ,

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

If n in NN , then find the value of i^n+i^(n+1)+i^(n+2)+i^(n+3) .

i ^n+i ^(n+1)+i ^(n+2)+i ^(n+3) (n∈N) is equal to