Example 3.16 Find the Fibonacci number $f(20)$.


Solution: Since $f(3)=3, f(4)=5, f(5)=8$, by Theorem 3.22 we get
$$
\begin{aligned}
f(20) & =f(10+10) \\
& =f(10) f(10)+f(9) f(9), \\
f(10) & =f(5+5)=f(5) f(5)+f(4) f(4) \\
& =8 \times 8+5 \times 5=89, \\
f(9) & =f(5+4)=f(5) f(4)+f(4) f(3) \\
& =8 \times 5+5 \times 3=55,
\end{aligned}
$$

Therefore,
$$
f(20)=89 \times 89+55 \times 55=7921+3025=10946 .
$$