I have to show the following:
Let $N_k=\frac{p_k}{q_k}$ with $\alpha=\langle 1;2,3,4,...,n,n+1\rangle$ and $n \in \mathbb{N}$. Then $\forall n \in \mathbb{N}$ with $n\geq 3$,
$p_n=n(p_{n-1}+p_{n-2})+(n-1)p_{n-3}+(n-2)p_{n-4}+\dots+3p_1+2p_0+2\;.$
$p_n$ is the n-th numerator of the convergent of $\alpha$ and $\alpha$ is a continued fraction, the number before the ";" is the part of the continued fraction in front of the "real" fraction, for example $\langle a;b,c\rangle =a+\frac{1}{b+\frac{1}{c}}$
As hint I got "Use Induction", but I don't know where to start.
I suggest:
$n=3 \rightarrow p(3)=3(p_2+p_1)+2p_0+2=3p_2+3p_1+2p_0+2$
Now I have to find out $p(3)$, so I have to get the continued fraction $\alpha=\langle 1;2,3\rangle$, but I don't know how to get my $\frac{p_k}{q_k}$.
Any hints would be helpful.