Question

Evaluate the sum (for math nerds)


`i {}^(0!) + i {}^(1!) + i {}^(2!) + i {}^(3!) + ... + i {}^(100!)`
Note that :

`i = \sqrt[]{ - 1}`
​

Answer 1

`n!` is divisible by 4 for all
`n\ge4`. This means, for instance,

`i^(4!) = \left(i^4\right)^(3!) = 1^(3!) = 1`

`i^(5!) = \left(i^4\right)^(5*3!) = 1^(5*3!) = 1`

etc, so that
`i^(n!) = 1` for all
`n\ge4`.

Meanwhile,

`i^(0!) = i^1 = i`

`i^(1!) = i^1 = i`

`i^(2!) = i^2 = -1`

`i^(3!) = i^6 = (-1)^3 = -1`

Then the sum we want is

`i^(0!) + i^(1!) + i^(2!) + i^(3!) + 97*1 = i + i - 1 - 1 + 97 = \boxed{95+2i}`

Answer 2

Answer: i+96

Explanation:

Note that
`i^(4k)`, where k is an integer, is equal to 1.

This means that
`i^(4!)=i^(5!)=i^(6)=\cdots=i^(99!)+i^(100!)=1`

So, we can rewrite the sum as
`i^(1)+i^(1)+i^(2)+i^3+97(1)=i+i-1-i+97=i+96`