It is currently Tue Jun 18, 2019 9:48 pm

 All times are UTC [ DST ]

 Page 1 of 1 [ 1 post ]
 Print view Previous topic | Next topic
Author Message
 Post subject: n-th root of a continued fraction of Lucas numbersPosted: Fri Jan 06, 2017 10:09 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
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.

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 1 [ 1 post ]

 All times are UTC [ DST ]

#### Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum

Search for:
 Jump to:  Select a forum ------------------ Algebra    Linear Algebra    Algebraic Structures    Homological Algebra Analysis    Real Analysis    Complex Analysis    Calculus    Multivariate Calculus    Functional Analysis    Measure and Integration Theory Geometry    Euclidean Geometry    Analytic Geometry    Projective Geometry, Solid Geometry    Differential Geometry Topology    General Topology    Algebraic Topology Category theory Algebraic Geometry Number theory Differential Equations    ODE    PDE Probability & Statistics Combinatorics General Mathematics Foundation Competitions Archives LaTeX    LaTeX & Mathjax    LaTeX code testings Meta