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Derive a formula for $C_k-C_{k-3}$ in terms of partial quotients $a_k$ and nominal denominators $q_k$.

Recall that

$$C_k=\frac{p_k}{q_k}$$

where

$$\begin{matrix} \begin{align} p_0&=a_0 & q_0&=1\\ p_1&=a_0a_1+1 & q_1&=a_1\\ p_k&=a_kp_{k-1}+p_{k-2} & q_k&=a_kq_{k-1}+q_{k-2} \end{align} \end{matrix}$$

This is what I did:

$$\begin{align} C_k-C_{k-3}&=\frac{p_k}{q_k}-\frac{p_{k-3}}{q_{k-3}}=\frac{p_kq_{k-3}-p_{k-3}q_k}{q_kq_{k-3}}=\frac{(a_kp_{k-1}+p_{k-2})q_{k-3}-p_{k-3}(a_kq_{k-1}+q_{k-2})}{q_kq_{k-3}}\\ &=\frac{a_k(p_{k-1}q_{k-3}-p_{k-3}q_{k-1})+p_{k-2}q_{k-3}-p_{k-3}q_{k-2}}{q_kq_{k-3}}\\ &=\frac{a_k(p_{k-1}q_{k-3}-p_{k-3}q_{k-1})+(-1)^{k-3}}{q_kq_{k-3}} \end{align}$$

This is because of a theorem which states that

$$p_kq_{k-1}-p_{k-1}q_k=(-1)^{k-1}$$

I get stuck here: I do not know how to get rid of the $p$'s so that the expression will be in terms of $a$'s and $q$'s alone.

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up vote 2 down vote accepted

Notice that you have proved $$p_kq_{k-3}-p_{k-3}q_k=a_k(p_{k-1}q_{k-3}-p_{k-3}q_{k-1})+(-1)^{k-3}$$ So you have replaced $p_k$ and $q_k$ with $p_{k-1}$ and $q_{k-1}$ (and introduced $a_k$, and $(-1)^{k-3}$). Perhaps if you do the process two more times you will get $p_{k-3}q_{k-3}-p_{k-3}q_{k-3}$, and your troubles will be over.

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