1B. The sequence $\left(a_{n}\right)_{n=1}^{\infty}$ is defined by the recursive formula

$$
a_{\circ}=-1, a_{n}=\frac{2 a_{n-1}-3}{3 a_{n-1}-4}, n \in \mathbb{N} .
$$

Determine the general term $a_{n}$ of the sequence.
Solution. If we substitute in the recursive formula, we get

$$
a_{1}=\frac{5}{7}, a_{2}=\frac{11}{13}, a_{3}=\frac{17}{19}, a_{4}=\frac{23}{25}, a_{5}=\frac{29}{31}
$$

or equivalently

$$
a_{1}=\frac{6 \cdot 1-1}{6 \cdot 1+1}, a_{2}=\frac{6 \cdot 2-1}{6 \cdot 2+1}, a_{3}=\frac{6 \cdot 3-1}{6 \cdot 3+1}, a_{4}=\frac{6 \cdot 4-1}{6 \cdot 4+1}, a_{5}=\frac{6 \cdot 5-1}{6 \cdot 5+1} .
$$

We will prove by mathematical induction that $a_{n}=\frac{6 n-1}{6 n+1}$.

Assume that the statement holds for $n=k$, i.e., $a_{k}=\frac{6 k-1}{6 k+1}$. Then for $n=k+1$ we have

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

By the principle of mathematical induction, we have that $a_{n}=\frac{6 n-1}{6 n+1}$ for every natural number.