Context:Problem involving the MGF of a die toss
$M_x(t)=\sum^6_{i=1}e^{tx}\cdot p(x)=e^{t}\cdot\frac{1}{6}+e^{2t}\cdot\frac{1}{6}+...e^{6t}\cdot\frac{1}{6}=\frac{1}{6}\cdot e^t \frac{e^{6t}-1}{e^t-1}$
I'm having some difficulty understanding with this is true. I do understand that you can factor out $\frac{1}{6}e^t$, but I don't understand why the remaining sum turns into $\frac{e^{6t}-1}{e^t-1}$.
I have found 2 explanations, one of them seems to fit this, and the other doesn't and I'm having difficulty understanding when it is one would use one vs. the other
Option 1 (doesn't work, the terms are backward): $\sum^{n-1}_{k=0}ar^k=a\cdot\frac{1-r^n}{1-r}$
Option 2 (does work): $\sum_{k=1}^nx^k=\frac{x^n-1}{x-1}$
While it would seem that I had found my answer in option 2, I guess I am confused because nearly every source I could find online cited option 1, while option 2 I could only find once. Is this indeed a legitimate series, can I assume that it is usable to solve my given problem?
Let's try to use the first formula.
Now, $a= \frac16 e^t, r = e^t, n = 6$.
$$\sum_{k=0}^5 ar^k = \frac16 e^t \cdot \frac{1-e^{6t}}{1-e^t}=\frac16 e^t \cdot \frac{e^{6t}-1}{e^t-1}$$
Notice that $\frac{c}{d}= \frac{-c}{-d}$.