Difference between revisions of "L(n+1)L(n-1)-L(n)^2=5(-1)^(n+1)"

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The following formula holds:
 
The following formula holds:
 
$$L(n+1)L(n-1)-L(n)^2=5(-1)^{n+1},$$
 
$$L(n+1)L(n-1)-L(n)^2=5(-1)^{n+1},$$
where $L(n)$ denotes a [[Lucas number]].
+
where $L(n)$ denotes a [[Lucas numbers|Lucas number]].
  
 
==Proof==
 
==Proof==
  
 
==References==
 
==References==
* {{PaperReference|A Primer on the Fibonacci Sequence Part I|1963|S.L. Basin|author2=V.E. Hoggatt, Jr.|prev=F(n+1)F(n-1)-F(n)^2=(-1)^n|next=findme}}  
+
* {{PaperReference|A Primer on the Fibonacci Sequence Part I|1963|S.L. Basin|author2=V.E. Hoggatt, Jr.|prev=F(n+1)F(n-1)-F(n)^2=(-1)^n|next=L(n)=F(n+1)+F(n-1)}}  
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

Latest revision as of 00:25, 25 May 2017

Theorem

The following formula holds: $$L(n+1)L(n-1)-L(n)^2=5(-1)^{n+1},$$ where $L(n)$ denotes a Lucas number.

Proof

References