The I's Have It

Solution 1: Cyclical nature of $i^n$

We note that the powers of $i$ is cyclical with a period of 4 as follows:

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

Similarly, the sum of the powers of $i$ is also cyclical as follows:

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

Since $2018 \bmod 4 = 2 \implies s_{2018}=\boxed{i-1}$ .

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Solution 2: Sum of Geometric Progression

$\begin{aligned} s_{2018} & = \sum_{k=1}^{2018} i^k \\ & = \frac {i(i^{2018}-1)}{i-1} \\ & = \frac {i(-2)}{i-1} \\ & = \frac {2i(1+i)}{(1-i)(1+i)} \\ & = \boxed {i-1} \end{aligned}$

$i=i, i^2=-1, i^3=-i, i^4=1 \Rightarrow i+i^2+i^3+i^4=i-1-i+1=0$

Since $i^{4n+1}=i$ and $4 | 2016$ ,

$i+i^2+i^3+i^4+\ldots+i^{2016} = 0$

$i^{2017}+i^{2018}=\boxed{i-1}$