Let $\dfrac{A_{n}}{B_{n}}$ be the $n^{th}$ convergent (approximant) $ \frac{A_{n}}{B_{n}}=b_{0}+\dfrac{a_{1}}{b_{1}+\dfrac{a_{2}}{b_{2}+\dfrac{a_{3}}{\begin{array}{c} b_{3}+ \\ \\ \end{array} \begin{array}{cc} \ddots & \\ & \end{array} +\dfrac{a_{n-1}}{b_{n-1}+\dfrac{a_{n}}{b_{n}}}}}} $ of a continued fraction. $A_{n}$, $A_{n-1}$, and $A_{n-2}$ satisfy the recurrence $ \begin{eqnarray*} \begin{pmatrix} A_{n} \\ B_{n} \end{pmatrix} &=& \begin{pmatrix} A_{n-1} & A_{n-2} \\ B_{n-1} & B_{n-2} \end{pmatrix} \begin{pmatrix} b_{n} \\ a_{n} \end{pmatrix} \quad \text{for }n\geq 1, \\ && \\ \text{and } \begin{pmatrix} A_{-1} & A_{0} \\ B_{-1} & B_{0} \end{pmatrix} &=& \begin{pmatrix} 1 & b_{0} \\ 0 & 1 \end{pmatrix} , \end{eqnarray*} $ which may be proved by induction.
Question: Is there a non inductive proof?
EDIT: I changed the title. Based on the comment below by anon
The numerators and denominators are defined inductively - and I don't believe you'll find an explicit formula for the n-th convergent of an arbitrary continued fraction (though maybe of particular ones with specific patterns, like $\varphi$) - so I don't see any other way to say prove anything about the convergents except inductively.
I reformulate the question to
Question: Is it right that the fundamental recurrence of an arbitrary continued fraction cannot be proved without using mathematical induction?