easy with trick

The value of $i^n$ is cyclical, $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , $i^4=1$ , $i^5=i$ , $-1, -i, 1 , i...$ .

Let $k$ be the remainder of $n\div 4$ , then we have:

$i^n=\begin{cases} \space \space 1,\quad if\space k=0 \\ \space \space i,\quad if\space k=1 \\ -1,\quad if\space k=2 \\ -i, \quad if\space k=3 \end{cases}$ .

Since both $222024$ and $32$ are both divisible by $4$ ( $k=0$ ), therefore, $i^{222024}+i^{32}=1+1=\boxed{2}$ .

$i^{222024}+i^{32}=(i^{4})^{55506}+(i^{4})^{8}=1^{8}+1^{55506}=1+1=2$ .