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

Find the value of 1+i^2+i^4+i^6++i^(2n)

Show that: {i^(19)+(1/i)^(25)}^2=-4 (ii) {i^(17)-(1/i)^(34)}^2=2i (iii) {i^(18)+(1/i)^(24)}^3=0 (iv) i^n+i^(n+1)+i^(n+2)+i^(n+3)=0 for all n in Ndot

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

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

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

Find the value of n if: (i) (n+2)! = 12n! (ii) (n+2)! = 60(n-1)! (iii) (n+3)! = 2550 (n+1)! (iv) (n-2)! = 132. (n-4)! .

If I _(n)=int _(0)^(pi) (sin (2nx))/(sin 2x)dx, then the value of I _( n +(1)/(2)) is equal to (n in I) :

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

Find the least positive value of n it ((1+i)/(1-i))^(n)=1

Find the value of sumsum_(0leilejlt=n)c_i^n"" c_j^ndot