Imaginary game

Note that the value of $i^k$ is cyclical with a period of 4. That is

$i^k = \begin{cases} 1 & \text{if }k \bmod 4 = 0 \\ i & \text{if }k \bmod 4 = 1 \\ -1 & \text{if }k \bmod 4 = 2 \\ -i & \text{if }k \bmod 4 = 3 \end{cases}$

As a result, its sum is also cyclical as follows.

$\displaystyle \sum_{k=1}^n i^k = \begin{cases} 0 & \text{if }n \bmod 4 = 0 \\ i & \text{if }n \bmod 4 = 1 \\ i -1 & \text{if }n \bmod 4 = 2 \\ -1 & \text{if }k \bmod 4 = 3 \end{cases}$

Since $20 \bmod 4=0$ , $\displaystyle \sum_{k=1}^{20} i^k = \boxed{0}$ .

Note that $i^{4n+1} = i \ , \ i^{4n+2} = -1 \ , \ i^{4n+3} = -i \ , \ i^{4n} = 1 \ \text{for} n= 0,1,2, \cdots \\ (i^{4n+1}+i^{4n+2}+i^{4n+3}+ i^{4n}) = (i-1-i+1) = 0 \\ (i+i^2+i^3+i^4)+(i^5+i^6+i^7+i^8)+ \ldots +(i^{17}+i^{18}+i^{19}+i^{20})= 0+0+ \cdots +0 =0$