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The continued fraction where $a_0=1, a_1=a_2=\cdots=a_k=2$. Show that

$q_k=p_{k-1}+q_{k-1}$ and $p_k=2q_k-p_{k-1},$

given that $$p_0=a_0, p_1=a_0a_1+1,p_k=a_kp_{k-1}+p_{k-2}$$ and

$$q_0=1,q_1=a_1,q_k=a_kq_{k-1}+q_{k-2}.$$

I'm not too sure where to go about this one. Thanks for looking.

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    $\begingroup$ Why not use induction to prove the above statement? $\endgroup$ – freak_warrior Mar 10 '15 at 5:34

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