Question

Evaluate the sum (for math nerds)

Answer 1

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

$i^{4!} = \left(i^4\right)^{3!} = 1^{3!} = 1$

$i^{5!} = \left(i^4\right)^{5\times3!} = 1^{5\times3!} = 1$

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

$i^{0!} + i^{1!} + i^{2!} + i^{3!} + 97\times1 = i + i - 1 - 1 + 97 = \boxed{95+2i}$

Answer 2

Answer: i+96

Step-by-step 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 $i^{1}+i^{1}+i^{2}+i^3+97(1)=i+i-1-i+97=i+96$