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Let  \(A_0 = 6 \\ A_1 = 5 \\ A_n = A_{n - 1} + A_{n - 2} \text{for } n \geq 2\).

 

There is a unique ordered pair (c,d) such that \(c\phi^n + d\widehat{\phi}^n\) is the closed form for sequence A_n. Find c using the Fibonacci and Lucas number sequences.

 

\(\phi = \frac{1 + \sqrt{5}}{2} \; \text{and} \; \widehat{\phi} = \frac{1 - \sqrt{5}}{2}\)

 

A0=6, A1 = 5, A2=11, A3=16, A4 = 27, A5 = 43...

Im not seeing any patterns

 

pls help

 

I found this https://math.stackexchange.com/questions/2215103/lucas-and-fibonacci-numbers but i don't get the final solution  (whats c?)

 Dec 14, 2020
edited by Guest  Dec 14, 2020
 #1
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c = 3 - sqrt(5) and d = 3 + sqrt(5)

 Dec 15, 2020
 #2
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Your sequence from the 2nd term and up:

 

11, 16, 27, 43, 70, 113, 183, 296, 479, 775, 1254, 2029, 3283, 5312, 8595, 13907, ...etc.

 

The "closed form" from the 2nd term and up is:

 

a_n = 1/2 (17 F_n + 5 L_n), where F_n==Fibonacci number and L_n==Lucas number.

 Dec 15, 2020
 #3
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so what is c?

Guest Dec 16, 2020

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