Runtime blow-up of recursive Fibonacci

A plain recursive routine for the Fibonacci numbers, $F(n)=F(n-1)+F(n-2)$, needs $100$ seconds to compute $F(n)$ for some large $n$. Roughly how long will it take for $F(n+1)$, and how could you compute Fibonacci numbers faster?

About $162$ seconds. The run time $T(n)=T(n-1)+T(n-2)+c$ grows like $F(n)$, i.e. by the golden ratio $\varphi\approx1.618$ per step, so $100\varphi\approx162$ s. Iterating from the bottom up (keeping the last two values) takes $O(n)$, and matrix powering or fast doubling takes $O(\log n)$ operations.