+0

0
121
1
+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
+121003
+3

Here's   "  a  "

L(n)   = (Phi)^n  + (  -phi)^n

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

Apr 10, 2021