Geometric series sum $\sum_2^\infty e^{3-2n}$

$$\sum_2^\infty e^{3-2n}$$

The formulas for these things are so ambiguous I really have no clue on how to use them.

$$\frac {cr^M}{1-r}$$

$$\frac {1e^2}{1-\frac{1}{e}}$$

Is that a wrong application of the formula and why?

Note $e^{3-2n}=e^3(e^{-2})^n$ so $$\sum_{n=2}^\infty e^{3-2n}=e^3\sum_{n=2}^\infty (e^{-2})^n=e^3\left(\sum_{n=0}^\infty(e^{-2})^n-1-e^{-2}\right).$$ You can take it from here.

The specific formula you want is $$\sum_{i=0}^\infty x^i = \frac{1}{1-x}.$$ There is no ambiguity there. But to apply this formula to the problem at hand, you first need to translate it so that you are summing from 0 to $\infty$, not from 2 to $\infty$. In this case, you can obtain $$\sum_{i=2}^\infty e^{3-2i} = e^3\sum_{i=0}^\infty e^{-2(i+2)}$$ from the more usefully-shaped $$e^3\sum_{i=0}^\infty e^{-2i}$$ by multiplying every term by $e^{-4}$. This is a common trick for geometric series - by just multiplying by powers of your radix, you can change which values you are summing over.

From there it seems like you've got the right idea, though your value for $r$ is slightly wrong. (Maybe this was a typo?)