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65
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avatar+79 

Part a:

The Fibonacci sequence, with \(F_0 = 0\)\(F_1 = 1\) and \(F_n = F_{n - 2} + F_{n - 1}\), has a closed form

 

\(F_n = \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right)\)

 

where

 

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

 

The Lucas numbers are defined in the same way, but with different starting values. Let \(L_0\) be the zeroth Lucas number and \(L_1\) be the first. If

 

\(\begin{align*} L_0 &= 2 \\ L_1 &= 1 \\ L_n &= L_{n - 1} + L_{n - 2} \; \text{for}\; n \geq 2 \end{align*}\)

 

then what is the tenth Lucas number? (Note: We seek a numerical answer.)

 

Part b:

Find (a, b) such that

 

\(L_n = a\phi^n + b\widehat{\phi}^n.\)
 

Part c:

Let

 

\(\begin{align*} A_0 &= 6 \\ A_1 &= 5 \\ A_n &= A_{n - 1} + A_{n - 2} \; \textrm{for} \; n \geq 2. \end{align*}\)


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.

 

Sorry for the long problem, but I really don't know how to solve it.

 Apr 10, 2021
 #1
avatar+118665 
+3

Here's   "  a  "

 

L(n)   = (Phi)^n  + (  -phi)^n   

 

L(10)   = [  ( 1 +sqrt 5)  /2) ]^10  +   [ ( sqrt 5  -  1)  /  2  ]^10   =    123

 

 

cool cool cool

 Apr 10, 2021

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