Consider the following continued fraction,
\begin{equation} \frac{3\left(\frac{3n+1}{2^{a_{1}}}\right)+1}{2^{a_{2}}}... \end{equation}
I am struggling to prove that the above is equivalent to \begin{equation} \frac{3^{m}n+3^{m-1}+\sum_{j=1}^{m-1}3^{m-1-j}2^{\sum_{\kappa=1}^{j}a_{\kappa}}}{2^{\sum_{i=1}^{m}a_{i}}}, \end{equation}
where $m$ is the number of occurrences of $a$ in the fraction. The proof is elementary for a constant $a$, but I can't find a method for generalised values. Could induction work here? Any help would be much appreciated.
If I understand you correctly, your continued fraction can be defined recursively like this: $$x_1:=\frac{3n+1}{2^{a_1}},\qquad x_{m+1}:=\frac{3x_m+1}{2^{a_{m+1}}}$$ If so, your claim can easily be proven by induction. Observe that your formula works for $m=1$. Also, let's set $A_j:=\sum_{k=0}^ja_k$ to make the calculations a little cleaner. Then for the induction step, \begin{align*}x_{m+1}&=\frac1{2^{a_{m+1}}}\left(3\cdot\frac{3^mn+3^{m-1}+\sum_{j=1}^{m-1}3^{m-1-j}2^{A_j}}{2^{A_m}}+1\right)\\&=\frac{3^{m+1}n+3^m+\sum_{j=1}^{m-1}3^{m-j}2^{A_j}}{2^{A_{m+1}}}+\frac{3^{m-m}2^{A_{m}}}{2^{A_{m+1}}}\\&=\frac{3^{m+1}n+3^m+\sum_{j=1}^m3^{m-j}2^{A_j}}{2^{A_{m+1}}}\end{align*} as claimed.