College Algebra, Chapter 9, 9.1, Section 9.1, Problem 70

Define the sequence
$\displaystyle G_n = \frac{1}{\sqrt{5}} \left( \frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{2^n} \right)$
Find the first 10 terms of this sequence using a calculator. Compare to the Fibonacci Sequence $F_n$

$
\begin{equation}
\begin{aligned}
G_ 1 = \frac{1}{\sqrt{5}} \left( \frac{(1 + \sqrt{5})^1 - (1 - \sqrt{5})^1}{2^1} \right) = 1\\
\\
G_ 2 = \frac{1}{\sqrt{5}} \left( \frac{(1 + \sqrt{5})^2 - (1 - \sqrt{5})^2}{2^2} \right) = 1\\
\\
G_ 3 = \frac{1}{\sqrt{5}} \left( \frac{(1 + \sqrt{5})^3 - (1 - \sqrt{5})^3}{2^3} \right) = 2\\
\\
G_ 4 = \frac{1}{\sqrt{5}} \left( \frac{(1 + \sqrt{5})^4 - (1 - \sqrt{5})^4}{2^4} \right) = 3\\
\\
G_ 5 = \frac{1}{\sqrt{5}} \left( \frac{(1 + \sqrt{5})^5 - (1 - \sqrt{5})^5}{2^5} \right) = 5\\
\\
G_ 6 = \frac{1}{\sqrt{5}} \left( \frac{(1 + \sqrt{5})^6 - (1 - \sqrt{5})^6}{2^6} \right) = 8\\
\\
G_ 7 = \frac{1}{\sqrt{5}} \left( \frac{(1 + \sqrt{5})^7 - (1 - \sqrt{5})^7}{2^7} \right) = 13\\
\\
G_ 8 = \frac{1}{\sqrt{5}} \left( \frac{(1 + \sqrt{5})^8 - (1 - \sqrt{5})^8}{2^8} \right) = 21\\
\\
G_ 9 = \frac{1}{\sqrt{5}} \left( \frac{(1 + \sqrt{5})^9 - (1 - \sqrt{5})^9}{2^9} \right) = 34\\
\\
G_{10} = \frac{1}{\sqrt{5}} \left( \frac{(1 + \sqrt{5})^{10} - (1 - \sqrt{5})^{10}}{2^{10}} \right) = 55\\
\end{aligned}
\end{equation}
$

Using Fibonacci Sequence $F_n = F_{n-1} + F_{n -2}$
Since $F_1 = 1$ and $F_2 = 2$, then

$
\begin{equation}
\begin{aligned}
F_3 &= F_2 + F_1 = 1 + 1 = 2 &&& F_7 &= F_6 + F_5 = 8 + 5 = 13\\
\\
F_4 &= F_3 + F_2 = 2 + 1 = 3 &&& F_8 &= F_7 + F_6 = 13 + 8 = 21\\
\\
F_5 &= F_4 + F_3 = 3 + 2 = 5 &&& F_9 &= F_8 + F_7 = 21 + 13 = 34\\
\\
F_6 &= F_5 + F_4 = 5 + 3 = 8 &&& F_{10} &= F_9 + F_8 = 34 + 21 = 55
\end{aligned}
\end{equation}
$