Find the value of `1+i^(2)+i^(4)+i^(6)+...+i^(2n),`
where`i=sqrt(-1)` and n in N.

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

The value of sum_(n=1)^(13) (i^n+i^(n+1)) , where i =sqrt(-1) equals (A) i (B) i-1 (C) -i (D) 0

Evaluate 1+i^2+i^4+i^6+...+i^(2n)dot

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

If (1+i)^(2n)+(1-i)^(2n)=-2^(n+1)(where,i=sqrt(-1) for all those n, which are

The value of i^2 + i^4 + i^6 + i^8.... upto (2n+1) terms , where i^2 = -1, is equal to:

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

Find the least positive integral value of n, for which ((1-i)/(1+i))^n , where i=sqrt(-1), is purely imaginary with positive imaginary part.

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