Not a typical geometric series

So, we have $\color{#EC7300}{z^7=1}$ . Let us try to write what-we-want in terms of $\color{#EC7300}{z^7}$ .

Let $\color{#69047E}{S=z^2+Z^4+Z^6+Z^8+...+z^{40}}$ . Then, multiplying both sides with $\color{#EC7300}{z^2}$ , (which we know is not equal to zero), we get:-

$\color{#69047E}{z^2S=+Z^4+Z^6+Z^8+z^{10}+...+z^{40}+z^{42}}$ .

Subtracting the two equations, we get:-

$\color{#69047E}{S(z^2-1)=z^{42}-z^2}$

Now, substituting $\color{#EC7300}{x^{42}=(x^{7})^6=1^6=1}$ , we get $\color{#69047E}{S=\frac{1-z^2}{z^2-1}}$ , so, $\boxed {S=-1}$

I love the combination of $\color{#EC7300}{orange}$ and $\color{#69047E}{purple}$ , they're just made for each other!

Series $z^2 + z^4$ and so on can be written as $2 + 3(z+z^2+z^3+z^4+z^5+z^6)$ (Try mod 7 for each power).

It can be further written as $2 + 3z(z^6-1)/(z-1)$ . $z^6$ can be written as $z^{-1}$ . Substituting value will give us $2 + 3(1-z)/(z-1)$
It is equal to 2-3=-1. Q.E.D If anyone can latex it, I will be thankful as I personally don't know how to use it.