A Golden (Formal) Power Series

As you probably know, I wear my heart on my sleeve:
Well, I took the golden opportunity (ha!) to bring the golden ratio $\Phi = \frac{1+\sqrt{5}}{2}$ into Calc 2 this week, using it (and its little pal $\Psi = \frac{1-\sqrt{5}}{2}$) to find a closed formula for the $n$-th term of the Fibonacci sequence.
The ubiquitous Fibonacci sequence! It’s something you may have encountered out in the wild. You know, it goes a little like this:
$$F_0 = 1, F_1 = 1, F_n = F_{n-1} + F_{n-2},$$
so that $$F_2 = 2, F_3 = 3, F_4 = 5, F_5 = 8, F_6 = 13, F_7 = 21, \ldots$$
And let’s say for some reason, you need to cook up $F_{108}$. I hope you have some time on your hands if you’re planning to add all the way up to that. Instead, wouldn’t it be nice if we had a simple formula that we could use — i.e., a formula that was not recursive — to figure out the $n$-th Fibonacci number?
Luckily, such a formula exists, and there are lots of ways to find it. In this post, we’ll find it using power series. Read on, brave blogosphere traveler.