How to compute $(i^2-i^4+i^6-i^8+...+i^{38})^2$

Question

How can i compute $(i^2-i^4+i^6-i^8+...+i^{38})^2$ ?

I can see that the powers are arithmetic progression with $d=2$ but i tried to compute

$S_{19}$ but it didn't work.

Thanks.

Answer 1

Note that $i^2=-i^4=i^6=-i^8=\ldots$ so that the sum simplifies significantly.

Answer 2

$\textbf{Hint:}$ Note that $i^{4n}-i^{4n+2}=2$ for $n \in \mathbb{N}$.

Answer 3

So we know that $i^2$ = -1, and that squared is $i^4$ which is 1. As aforementioned, $i^2 = -i^4$. Because, $i^4=1$, $i^2=i^6=i^{10}...i^{38}$, and $i^4=i^8..i^{36}$. Therefore, canceling out terms leaves us with $(i^{38})^2$, which evaluates to $(-1)^2$, which we know to be one.

==> $(((i^2-i^4)+(i^6-i^8)...(i^{34}-i^{36})+i^{38})^2$ = $(0+0+0+..+(-1))^2$ = (-1)^2 = 1