n-th root of a continued fraction of Lucas numbers
- Tolaso J Kos
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n-th root of a continued fraction of Lucas numbers
Let us denote the $n$ -th Lucas number as $L_n$. It is known that $L_0=2$ , $L_1=1$ as well as
$$L_{n+2}=L_{n+1} + L_{n} \quad \text{forall} \; n \geq 0$$
Calculate the value of
$$\mathcal{R}=\sqrt[n]{L_{2n} - \frac{1}{L_{2n}- \dfrac{1}{L_{2n}-\ddots}}}$$
I don't have a solution.
$$L_{n+2}=L_{n+1} + L_{n} \quad \text{forall} \; n \geq 0$$
Calculate the value of
$$\mathcal{R}=\sqrt[n]{L_{2n} - \frac{1}{L_{2n}- \dfrac{1}{L_{2n}-\ddots}}}$$
I don't have a solution.
Imagination is much more important than knowledge.
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