## Task B-2.1.

For which $n \in \mathbb{N}$ does the following hold

$$
1 i^{1}+2 i^{2}+3 i^{3}+4 i^{4}+\cdots+n i^{n}=48+49 i, \text { if } i=\sqrt{-1} ?
$$
## 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 $n = 97$.