Is there a formal way for solving these series with |r| < 1 $$\sum _{n=1}^{\infty }\:\left(\frac{5}{6}\right)^{2n-2}=\sum \:_{n=1}^{\infty \:}\:\frac{\left(\frac{5}{6}\right)^{2n}}{\left(\frac{5}{6}\right)^2}=\frac{1}{\left(\frac{5}{6}\right)^2}\sum \:\:_{n=1}^{\infty \:\:}\:\left(\frac{5}{6}\right)^{2n}=\frac{1}{\left(\frac{5}{6}\right)^2}\cdot \frac{\left(\frac{5}{6}\right)^2-\left(\frac{5}{6}\right)^{2n+1}}{1-\left(\frac{5}{6}\right)^2}=\frac{1}{\left(\frac{5}{6}\right)^2}\cdot \:\frac{\left(\frac{5}{6}\right)^2}{1-\left(\frac{5}{6}\right)^2}=\:\frac{1}{1-\left(\frac{5}{6}\right)^2}$$
I'm applying directly the formula:
$$\frac{a-a^{n+1}}{1-a}$$ As I have as exponen $$2n$$, then I just write:
$$\frac{a^2-a^{2n+1}}{1-a^2}$$
Is there any other formal way to do this? I know that 
$$\sum _{i=0}^{\infty }\:x^{2i}\:=\:\frac{1}{1-x^2}$$
But I don't understand what happened with -2 part of the exponencial 2n−2, or even how to generalize for $$\sum _{i=1,2,3...}^{\infty }$$
I found an idea here https://en.wikipedia.org/wiki/Geometric_progression and I guess that I could do:
$$\sum _{n=m}^{\infty }\:ar^{cn-b}=:\frac{ar^{m}}{(1-r^c)r^b}   ...or...   \sum _{n=m}^{\infty }\:ar^{cn+b}=:\frac{ar^{m+b}}{1-r^c} $$
Is there any other more simple way to calculate this?
Thanks.

The formal way of solving this problem is to map it to a geometric progression:
$$ \sum_{n=m}^{\infty} ar^{cn-b}=\sum_{n-m=0}^{\infty} ar^{-b} r^{cm}r^{c(n-m)} = ar^{-b} r^{cm} \sum_{i=0}^{\infty}\left( r^c\right)^{i} = \frac{ar^{cm-b}}{1-r^c} .$$
Here, we first use a change of variable $(n,m)\to(i=n-m,m)$. Then, we recognize that $\sum_{i=0}^{\infty}\left( r^c\right)^{i}$ is a geometric progression with the common ratio as $r^c$.
