Problem 797

## First Solution.

The sum of every four consecutive addends from the left side equals $2-2i$. 1 point

By summing the first 96 addends, we get $24 \cdot(2-2i)=48-48i$. 1 point

Therefore, for the initial equality to hold, there must be at least one more power of $i$ on the left side. Regardless of how many there are, we can write

$$\begin{aligned} & 24 \cdot(2-2i) + a + bi = 48 + 49i \\ & 48 + a + (-48 + b)i = 48 + 49i \end{aligned}$$

1 point

From this, we have

$$\begin{array}{ll} 48 + a = 48 \Rightarrow a = 0, & 1 \text{ point} \\ -48 + b = 49 \Rightarrow b = 97. & 1 \text{ point} \end{array}$$

We conclude that there are 97 addends on the left side.

The 97th addend is $97i = 97i^{97}$, so the desired $n = 97$.