The Fibonacci sequence is defined recursively by $F_{n+2}=F_{n+1}+F_{n}$ for $n \in \mathbb{Z}$ and $F_{1}=F_{2}=1$. Determine the value of:

$$
\left(1-\frac{F_{2}^{2}}{F_{3}^{2}}\right)\left(1-\frac{F_{3}^{2}}{F_{4}^{2}}\right) \cdot \ldots \cdot\left(1-\frac{F_{2019}^{2}}{F_{2020}^{2}}\right)
$$
Solution

We can rewrite a generic term of the given product as

$$
\begin{aligned}
1-\frac{F_{i}^{2}}{F_{i+1}^{2}} & =\frac{F_{i+1}^{2}-F_{i}^{2}}{F_{i+1}^{2}} \\
& =\frac{\left(F_{i+1}-F_{i}\right)\left(F_{i+1}+F_{i}\right)}{F_{i+1}^{2}} \\
& =\frac{F_{i-1} F_{i+2}}{F_{i+1}^{2}}
\end{aligned}
$$

Therefore,

$$
\begin{aligned}
\left(1-\frac{F_{2}^{2}}{F_{3}^{2}}\right)\left(1-\frac{F_{3}^{2}}{F_{4}^{2}}\right)\left(1-\frac{F_{4}^{2}}{F_{5}^{2}}\right) \ldots \cdot\left(1-\frac{F_{2018}^{2}}{F_{2019}^{2}}\right) \cdot\left(1-\frac{F_{2019}^{2}}{F_{2020}^{2}}\right) & = \\
\frac{F_{1} F_{4}}{F_{3}^{2}} \cdot \frac{F_{2} F_{5}}{F_{4}^{2}} \cdot \frac{F_{3} F_{6}}{F_{5}^{2}} \cdot \ldots \cdot \frac{F_{2017} F_{2020}}{F_{2019}^{2}} \cdot \frac{F_{2018} F_{2021}}{F_{2020}^{2}} & = \\
\frac{F_{1} F_{2} \cdot\left(F_{3} F_{4} \ldots F_{2017} F_{2018}\right) \cdot F_{2021}}{F_{3} \cdot\left(F_{3} F_{4} \ldots F_{2018} F_{2019}\right) F_{2019} F_{2020}} & = \\
\frac{F_{2021}}{2 F_{2019} F_{2020}} & =
\end{aligned}
$$