+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.
