Closed formula for Lucas numbers

Asked Apr 20 '12 at 05:59

Active May 04 '15 at 20:44

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Possible Duplicate:
Prove this formula for the Fibonacci Sequence
How does one find a formula for the recurrence relation $a_{1}=1,a_{2}=3, a_{n+2}=a_{n+1}+a_{n}?$

How do I go about obtaining a closed formula for Lucas numbers?

The Lucas numbers $ L_n $ are defined by $L_1 = 1$, $L_2 = 3$, and $L_n = L_{n-1} + L_{n-2}$.

I tried looking into it on my discrete math textbook but I'm really confused. Maybe someone can lay out the steps for it?

edited Apr 13 '17 at 12:19

Community

asked Apr 20 '12 at 05:59

soniccool

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1

http://en.wikipedia.org/wiki/Lucas_numbers#Relationship_to_Fibonacci_numbers – Pedja Apr 20 '12 at 06:02
These are two different questions. – soniccool Apr 20 '12 at 06:03
You can use the techniques in this question. – J. M. ain't a mathematician Apr 20 '12 at 06:04
Part (b) of the question Gerry linked is precisely the same as this one. – anon Apr 20 '12 at 06:12
b) is a bit different that one has a different value at ='s though. – soniccool Apr 20 '12 at 06:14
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mystycs: The questions are the same. The conditions $L_{n+2}=L_{n+1}+L_n$ and $L_n=L_{n-1} + L_{n-2}$ (with appropriate range for $n$'s$) are equivalent. If you still think that there is a difference, could you explain what difference is there? – Martin Sleziak Apr 20 '12 at 06:28
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The beauty of mathematics is that if you know that 2 apples plus 2 apples is 4 apples, then even though 2 oranges plus 2 oranges is a different question, you can apply what you have learned to solve it. So it is with the current question and the earlier one. – Gerry Myerson Apr 20 '12 at 06:42

Closed formula for Lucas numbers

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