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: HINTS: 
$$i+i^2+i^3+i^4=0,$$
$$i^{4m+n}=i^{n}.$$
A: One trick for this kind of problem is to work it out for small numbers and see if you can find a pattern. Try working out:
$$ i + i^2 $$
$$ i + i^2 + i^3 $$
$$ i + i^2 + i^3 + i^4 $$
$$ i + i^2 + i^3 + i^4 + i^5 $$
$$ i + i^2 + i^3 + i^4 + i^5 + i^6$$
Do you see a pattern?
A: $$a + a^2 + a^3 + \dots a^n = a\frac{1-a^{n}}{1-a}$$
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 $-2i$.
A: Your first idea works:
If $n=4k+1$ then $i^n = i$,
If $n=4k+2$ then $i^n = -1$,
If $n=4k+3$ then $i^n = -i$,
If $n=4k$ then $i^n = 1$
So,
$i+i^2+i^3+i^4=0$
$i^5+i^6+i^7+i^8=0$
.
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$i^{97}+i^{98}+i^{99}+i^{100}=0$
Then, in the sum $i+i^2+i^3+...+i^{102}$ the only left terms are $i^{101}+i^{102}=i-1$
Thus, $(i+i^2+i^3+...+i^{102})^2=(i-1)^2= i^2 -2i+1=-1-2i+1=-2i$
A: $$i=i\\i^2 =-1\\i^3 =-i \\i^4 =1$$
Thus, $$\sum_{k=1}^{4n} i^k =0\  \text{for positive integer}\  n$$,
Hence, $$\sum_{k=1}^{102} i^k=\sum_{k=1}^{2} i^k=-1+i$$
And it's square equals $-2i$
