Solve: $(i+i^2+i^3+...+i^{102})^2$ 
Solve $$(i+i^2+i^3+...+i^{102})^2$$

I have tried 2 approaches: 


*

*to look at power of $2 (=-1)$ and power of $4 (=1)$ and there are $102$ elements which is even number, but because there are $4$ "types" $\{i,-1,-i,1\}$ it did not work it out

*to look at 
$$(i+i^2+i^3+...+i^{102})^2=\left(\sum_{k=1}^{102}i^k\right)^2$$ but $$\left(\sum_{k=1}^{102}i^k\right)^2\neq \sum_{k=1}^{102}i^{2k}$$ so it was not useful.


Any suggestion how to tackle this?

$$a + a^2 + a^3 + \dots a^n = a\frac{1-a^{n}}{1-a}$$
Therefore,
\begin{align*}
(i+i^2 + \cdots + i^{102}) &= i \frac{1-i^{102}}{1-i}\\
& = i \frac{1-i^{100}i^2}{1-i} \\
 &= i \frac{1+(i^{4})^{25}}{1-i}\\
 &= \frac{2i}{1-i}\\
&= -1 + i
\end{align*}
Squaring yields $-2i$.
