Question

1+ i^2+i^3+i^4+⋯+i^2n evaluate :
(a) Positive
(b) negative
(c) 0
(d) can not be evaluate



please explain

Answer 1

(-There was an error in the answer. +I fixed it.)

If we look at i, i², i³, and 1 we can find a useful equation.

The sum of them is 0. ∴1+i+i²+i³=0

2n could be whether a multiple of 2 and 4.

We could pick any number between them and find the solutions.

Let 2 be the number. We get 1+i²=0. In this case, it's zero.

Let 4 be the number. We get (1+i+i²+i³)+1=0+1. In this case, it's 1.

All cases were tested and it is not one solution. Therefore it cannot be evaluated. (This is the concept of functions.)

Answer 2

Answer:

A

Step-by-step explanation:

1+i+i^2+i^3+........+i^2n

where n is natural numbers then n=1,2,3,4......

i= -1 then 1+(-1)+1+(-1)+...........+1=0

in this way

1+i^2+i^3+.......+i^2n

1+1+(-1)+1+(-1)+.........+1= 1

the it gets positive value