n-th root of a continued fraction of Lucas numbers

General Mathematics
Post Reply
User avatar
Tolaso J Kos
Administrator
Administrator
Posts: 867
Joined: Sat Nov 07, 2015 6:12 pm
Location: Larisa
Contact:

n-th root of a continued fraction of Lucas numbers

#1

Post by Tolaso J Kos »

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.
Imagination is much more important than knowledge.
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 6 guests