Chapter 12 Sequences and Series - 12.5 Use Recursive Rules with Sequences and Functions - 12.5 Exercises - Problem Solving - Page 832: 46

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Given: $f_n=\frac{1}{\sqrt 5}(\frac{1+\sqrt 5}{2})^n-\frac{1}{\sqrt 5}(\frac{1-\sqrt 5}{2})^n$ The first five terms are: $a_1=\frac{1}{\sqrt 5}(\frac{1+\sqrt 5}{2})^n-\frac{1}{\sqrt 5}(\frac{1-\sqrt 5}{2})^n\\=\frac{1}{\sqrt 5}(\frac{1+\sqrt 5}{2})^1-\frac{1}{\sqrt 5}(\frac{1-\sqrt 5}{2})^1\\=\frac{1}{\sqrt 5}(\sqrt 5)\\=1$ $a_2=\frac{1}{\sqrt 5}(\frac{1+\sqrt 5}{2})^n-\frac{1}{\sqrt 5}(\frac{1-\sqrt 5}{2})^n\\=\frac{1}{\sqrt 5}(\frac{1+\sqrt 5}{2})^2-\frac{1}{\sqrt 5}(\frac{1-\sqrt 5}{2})^2\\=\frac{1}{\sqrt 5}(\sqrt 5)\\=1$ $a_3=\frac{1}{\sqrt 5}(\frac{1+\sqrt 5}{2})^n-\frac{1}{\sqrt 5}(\frac{1-\sqrt 5}{2})^n\\=\frac{1}{\sqrt 5}(\frac{1+\sqrt 5}{2})^3-\frac{1}{\sqrt 5}(\frac{1-\sqrt 5}{2})^3\\=\frac{1}{\sqrt 5}(4.489856)\\=2$ $a_4=\frac{1}{\sqrt 5}(\frac{1+\sqrt 5}{2})^n-\frac{1}{\sqrt 5}(\frac{1-\sqrt 5}{2})^n\\=\frac{1}{\sqrt 5}(\frac{1+\sqrt 5}{2})^4-\frac{1}{\sqrt 5}(\frac{1-\sqrt 5}{2})^4\\=\frac{1}{\sqrt 5}(6.73971)\\=3$ $a_5=\frac{1}{\sqrt 5}(\frac{1+\sqrt 5}{2})^n-\frac{1}{\sqrt 5}(\frac{1-\sqrt 5}{2})^n\\=\frac{1}{\sqrt 5}(\frac{1+\sqrt 5}{2})^5-\frac{1}{\sqrt 5}(\frac{1-\sqrt 5}{2})^5\\=\frac{1}{\sqrt 5}(11.24932)\\=5$