It is possible to derive a general formula for $F_n$ without computing all the previous numbers in the sequence. If a gemetric series (i.e. a series with a constant ratio between consecutive terms $r^n$) is to solve the difference equation, we must have

\begin{aligned} r^n = r^{n-1} + r^{n-2} \\ \end{aligned}

which is equivalent to

\begin{aligned} r^2 = r + 1 \\ \end{aligned}

This equation has two unique solutions \begin{aligned} \varphi = & \frac{1 + \sqrt{5}}{2} \approx 1.61803\cdots \ \psi = & \frac{1 - \sqrt{5}}{2} = 1 - \varphi = - {1 \over \varphi} \approx -0.61803\cdots \ \end{aligned}

In particular the larger root is known as the golden ratio \begin{align} \varphi = \frac{1 + \sqrt{5}}{2} \approx 1.61803\cdots \end{align}

Now, since both roots solve the difference equation for Fibonacci numbers, any linear combination of the two sequences also solves it

\begin{aligned} a \left(\frac{1 + \sqrt{5}}{2}\right)^n + b \left(\frac{1 - \sqrt{5}}{2}\right)^n \\ \end{aligned}

It's not hard to see that all Fibonacci numbers must be of this general form because we can uniquely solve for $a$ and $b$ such that the initial conditions of $F_1 = 1$ and $F_0 = 0$ are met

\begin{equation} \left\{ \begin{aligned} F_0 = 0 = a \left(\frac{1 + \sqrt{5}}{2}\right)^0 + b \left(\frac{1 - \sqrt{5}}{2}\right)^0 \\ F_1 = 1 = a \left(\frac{1 + \sqrt{5}}{2}\right)^1 + b \left(\frac{1 - \sqrt{5}}{2}\right)^1 \\ \end{aligned} \right. \end{equation}

yielding

\begin{equation} \left\{ \begin{aligned} a = \frac{1}{\sqrt{5}} \\ b = \frac{-1}{\sqrt{5}} \\ \end{aligned} \right. \end{equation}

We have therefore derived the general formula for the $n$-th Fibonacci number

\begin{aligned} F_n = \frac{1}{\sqrt{5}} \left(\frac{1 + \sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}} \left(\frac{1 - \sqrt{5}}{2}\right)^n \\ \end{aligned}