3.2.25 * (Fibonacci sequence) Assuming a pair of rabbits produces a new pair (one male and one female) every month, and each new pair starts producing a new pair (one male and one female) every month after two months. At the beginning of the year, a pair of adult rabbits is placed in the rabbit house. How many pairs of rabbits will there be in the rabbit house after one year?
Let $f_{n}$ denote the number of pairs of rabbits in the rabbit house at the beginning of the $n$-th month. It is easy to see that $f_{1}=1, f_{2}=2, f_{3}=3$. At the beginning of the $(n+2)$-th month, the rabbits in the rabbit house can be divided into two parts: one part consists of the rabbits that were already in the rabbit house at the beginning of the $(n+1)$-th month, totaling $f_{n+1}$ pairs, and the other part consists of the newborn rabbits at the beginning of the $(n+2)$-th month, totaling $f_{n}$ pairs. Thus, $f_{n+2}=f_{n+1}+f_{n}$. The corresponding characteristic equation is $x^{2}=x+1$, with characteristic roots $x_{1}=\frac{1+\sqrt{5}}{2}, x_{2}=\frac{1-\sqrt{5}}{2}$. Therefore, $f_{n}=A_{1}\left(\frac{1+\sqrt{5}}{2}\right)^{n}+A_{2}\left(\frac{1-\sqrt{5}}{2}\right)^{n}$. Using the initial conditions $f_{1}=1, f_{2}=2$, we get
$$
\left\{\begin{array}{l}
\left(\frac{1+\sqrt{5}}{2}\right) A_{1}+\left(\frac{1-\sqrt{5}}{2}\right) A_{2}=1, \\
\left(\frac{1+\sqrt{5}}{2}\right)^{2} A_{1}+\left(\frac{1-\sqrt{5}}{2}\right)^{2} A_{2}=2,
\end{array}\right.
$$

Solving these, we find $A_{1}=\frac{1+\sqrt{5}}{2 \sqrt{5}}, A_{2}=-\frac{1-\sqrt{5}}{2 \sqrt{5}}$, so $f_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$.