The answer is -1.
let's first start with small numbers, i^1 + i^2 + i^3 + i^4. Assuming in your case i is the imaginary number, I'll start by calculating that. i^1 is just i. i^2 is -1, i^3 is -1(i) which is -i, and i^4 is -i(i) which is -i^2 which is -1(-1) which becomes 1. Continiuing to calculate these, you'll find that there's a pattern.
i^1 = i i^5 = i
i^2 = -1 i^6 = -1
i^3 = -i i^7 = -i
i^4 = 1 i^8 = 1
The powers of i repeat in groups of 4.
Now that we've found out the pattern, we'll calculate yours. 99/4 leaves a remainder of 3. So then we have i^97 = i, i^98 = -1, and i^99 = -1. The i and -i cancel out, so you're left with the answer -1.
First find the sum of the numbers.
Also, since 0 has no value, I'll take it out for now.
The number series 1, 2, 3, 4, . . . . , 98, 99.
The first term a = 1
The common difference d = 1
Total number of terms n = 99
apply the input parameter values in the formula
Sum = n/2 x (a + Tn)
= 99/2 x (1 + 99)
1 + 2 + 3 + 4 + . . . . + 98 + 99 = 4950
Therefore, 4950 is the sum of positive integers from 1 to 99.
Next find the average.
The amount of numbers = 100
4950/100 = 49.5
So your final answer is 49.5