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For $n>0$ and $a_1,...,a_n \in K$ let $c_n(a_1,...,a_n)$ be the determinant of the matrix

$$ \begin{pmatrix} a_1 & 1 & 0 & \cdots & 0 \\ -1 & a_2 & \ddots & \ddots & \vdots \\ 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & \ddots & 1 \\ 0 & \cdots & 0 & -1 & a_n \end{pmatrix} $$

Show for $n \ge 2$ that following holds:

$$ \frac{c_n(a_1,...,a_n)}{c_{n-1}(a_2,...,a_n)} = a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots + \cfrac{1}{a_{n-1}+\frac{1}{a_n}}}} $$

I want to show it using induction over n but I already fail at the initial step:

For $n = 2$ I have to show: $$ \frac{c_2(a_1,a_2)}{c_{2-1}(a_2)} = a_1 + \frac{1}{a_2} $$

which is $$ \frac{a_1a_2 + 1}{a_2} \neq a_1 + \frac{1}{a_2} $$

My also have no idea for the induction step where I have to show: $$ \frac{c_{n+1}(a_1,...,a_{n+1})}{c_n(a_2,...,a_{n+1})} = a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots + \cfrac{1}{a_n+\frac{1}{a_{n+1}}}}} $$

So first I calculate $c_{n+1}(...)$ by developing after the first column which gives me: $$ a_1 \cdot \det \begin{pmatrix} a_2 & 1 & & & & \\ -1 & a_3 & 1 & & & \\ & -1 & \ddots & & & \\ & & & \ddots & & \\ & & & & & 1 \\ & & & & -1 & a_{n+1} \end{pmatrix} + \det \begin{pmatrix} 1 & 0 & & & & \\ -1 & a_3 & 1 & & & \\ & -1 & a_4 & & & \\ & & & \ddots & & \\ & & & & & 0 \\ & & & & 0 & a_{n+1} \end{pmatrix} $$ which after developing after the first row gives me $$ a_1 \cdot c_n(a_2,...,a_{n+1}) + c_{n-2}(a_3,...,a_{n+1}). $$

Analog for $c_n(...)$: $$ c_n(...) = a_2 c_{n-1}(...) + c_{n-3}(...) $$

So I have $$ \frac{c_{n+1}(...)}{c_n(...)} = \frac{a_1c_n(...) + c_{n-2}(...)}{a_2c_{n-1}(...) + c_{n-3}(...)} $$

written in another way it is $$ \frac{a_1c_n(...)}{c_n(...)} + \frac{c_{n-2}}{a_2c_{n-1}(...) + c_{n-3}(...)} $$

I write it as $$ a_1 + \frac{1}{a_2 \frac{c_{n-1}(...)}{c_{n-2}(...)} + \frac{c_{n-3}(...)}{c_{n-2}(...)}} $$

and then $$ a_1 + \frac{1}{a_2 \frac{c_{n-1}(...)}{c_{n-2}(...)} + \frac{1}{\frac{c_{n-2}(...)}{c_{n-3}(...)}}} $$

so $$ a_1 + \frac{1}{a_2 \cdot \left( a_3 + \cfrac{1}{a_4 + \cfrac{1}{\ddots + \cfrac{1}{a_n+\frac{1}{a_{n+1}}}}} \right) + \frac{1}{\left( a_4 + \cfrac{1}{a_5 + \cfrac{1}{\ddots + \cfrac{1}{a_n+\frac{1}{a_{n+1}}}}} \right)}} $$

I dont know how to preced from here, any help or simpler solutions are appreciated!

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but I already fail at the initial step

Not really. You have

$$\frac{c_2(a_1,a_2)}{c_1(a_2)} = \frac{a_1a_2+1}{a_2} = \frac{a_1a_2}{a_2} + \frac{1}{a_2} = a_1 + \frac{1}{a_2}.$$

In your induction step, you also have the necessary ingredients,

$$c_{n+1}(a_1,\dotsc,a_{n+1}) = a_1\cdot c_n(a_2,\dotsc,a_{n+1}) + c_{n-1}(a_3,\dotsc,a_{n+1}),$$

you just have made a dimension error and thought it was $c_{n-2}(a_3,\dotsc,a_{n+1})$. The dimension of that matrix is $\bigl((n+1)-2\bigr)^2$, not $(n-2)^2$.

And then you rename, calling $b_k = a_{k+1}$ for $k = 1,\dotsc, n$, to get

$$\begin{align} \frac{c_{n+1}(a_1,\dotsc,a_{n+1})}{c_n(a_2,\dotsc,a_{n+1})} &= \frac{a_1\cdot c_n(a_2,\dotsc,a_{n+1}) + c_{n-1}(a_3,\dotsc,a_{n+1})}{c_n(a_2,\dotsc,a_{n+1})}\\ &= a_1 + \frac{c_{n-1}(a_3,\dotsc,a_{n+1})}{c_n(a_2,\dotsc,a_{n+1})}\\ &= a_1 + \frac{1}{\frac{c_n(b_1,\dotsc,b_n)}{c_{n-1}(b_2,\dotsc,b_n)}}\\ &= a_1 + \cfrac{1}{b_1 + \cfrac{1}{b_2 + \cfrac{1}{\ddots + \cfrac{1}{b_{n-1}+ \frac{1}{b_n}}}}}\\ &= a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{\ddots + \cfrac{1}{a_{n}+ \frac{1}{a_{n+1}}}}}} \end{align}$$

using the induction hypothesis on the sequence $b_1,\dotsc,b_n$.

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