Show $p(n)=n(p_{n-1}+p_{n-2})+(n-1)p_{n-3}+(n-2)p_{n-4}+...+3p_1+2p_0+2$ 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.
 A: To get you started:
The successive convergents of $\langle 1;2,3\rangle$ are $\langle 1\rangle = \frac11$, $\langle 1;2\rangle = \frac32$, and $\langle 1;2,3\rangle = 1+\frac1{2+\frac13}=1+\frac1{7/3}=$  $1+\frac37=\frac{10}3$, so $p_0=1$, $p_1=3$, and $p_2=10$. Thus, $3p_2+3p_1+2p_0+2=43$. And $\langle 1;2,3,4\rangle =$  $1+\frac1{2+\frac1{3+\frac14}}=1+\frac1{2+\frac4{13}}=1+\frac{13}{30}=\frac{43}{30}$, so $p_3$ is indeed $43$.
You might want to look at this material for information on how the convergents are calculated recursively.
Added: Your continued fraction is $\langle a_0;a_1,a_2,\dots\rangle$, where $a_n=n+1$. Thus, the recurrence given in the Wikipedia article becomes $p_n=(n+1)p_{n-1}+p_{n-2}$, or $p_{n+1}=(n+2)p_n+p_{n-1}$. Assume as an induction hypothesis that 
$$p_n=n(p_{n-1}+p_{n-2})+(n-1)p_{n-3}+(n-2)p_{n-4}+\dots+3p_1+2p_0+2$$ and 
$$p_{n-1}=(n-1)(p_{n-2}+p_{n-3})+(n-2)p_{n-4}+(n-3)p_{n-5}+\dots+3p_1+2p_0+2\;.$$ Then $$\begin{align*}
p_{n+1}&=(n+2)p_n+p_{n-1}\\
&=(n+1)p_n+p_n+p_{n-1}\\
&=(n+1)p_n+\left(np_{n-1}+\sum_{k=0}^{n-2}(k+2)p_k+2\right)+p_{n-1}\\
&=(n+1)(p_n+p_{n-1})+\sum_{k=0}^{n-2}(k+2)p_k+2\;,
\end{align*}$$
which is exactly what you need it to be for the induction to go through.
