2. The sequence $a_{n}$ is defined recursively as

$$
\begin{aligned}
& a_{0}=-1 \\
& a_{n}=\frac{2 a_{n-1}-3}{3 a_{n-1}-4}, \quad n \in \mathbf{N}
\end{aligned}
$$

Determine the formula for $a_{n}$.
2. Let's calculate the first few terms of this sequence:

$$
a_{1}=\frac{5}{7}, \quad a_{2}=\frac{11}{13}, \quad a_{3}=\frac{17}{19}
$$

Assume that the correct formula is

$$
a_{n}=\frac{6 n-1}{6 n+1}
$$

10 points

We will prove this by mathematical induction.

$1^{\circ} \text{For} n=1$ the formula holds.

$2^{\circ}$ Assume that it holds for some $n \geq 1$. Then

$$
a_{n+1}=\frac{2 \cdot \frac{6 n-1}{6 n+1}-3}{3 \cdot \frac{6 n-1}{6 n+1}-4}=\frac{6 n+5}{6 n+7}=\frac{6(n+1)-1}{6(n+1)+1}
$$

so the statement holds for $n+1$ as well. By the principle of mathematical induction, it holds for every $n \geq 1$.

15 points