If $\frac{p_r}{q_r}$ be the $r^{\text{th}}$ convergent of the continued fraction of $\frac{\sqrt{5}+1}{2}$ then prove that $p_{n+1}=p_{n}+p_{n-1}$ and $p_{2n}=p_{2n-1}+p_{2n-2}$.
Attempt:
I have written the continued fraction of $\frac{\sqrt{5}+1}{2}$ which is equal to $1+\frac{1}{1+}\frac{1}{1+}\frac{1}{1+}\cdots$. But how to get the above two relations. Please help.
It's quite standard
$\displaystyle 1=\frac{1}{1}=\frac{p_{0}}{q_{0}}$
$\displaystyle 1+\frac{1}{1}=\frac{2}{1}=\frac{p_{1}}{q_{1}}$
$\displaystyle 1+\frac{1}{\frac{p_{n}}{q_{n}}}=\frac{p_{n}+q_{n}}{p_{n}}=\frac{p_{n+1}}{q_{n+1}}$
Note that if $\gcd(p_{n},q_{n}) = 1 \implies \gcd(p_{n},p_{n}+q_{n}) = 1$
Equating numerators and denominators then eliminate $q_{n}$. The result follows at once.
Put $m=2n-1$, then $p_{m+1}=p_{m}+p_{m-1} \implies p_{2n}=p_{2n-1}+p_{2n-2}$