Chapter 9.3, Problem 50E

Solution

To identify: The first seven terms of the sequence whose $nth$ term How do these seven terms compare with the first seven terms of the Fibonacci

sequence?.

  $a_n=\frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^n-\frac{1}{\sqrt{5}}{\left(\frac{1-\sqrt{5}}{2}\right)}^n$

The term of the given sequence are the term of the Fibonacci sequence.

  $1,1,2,3,5,8,\mathrm{13...}$

Given information:

The given $nth$ term is

  $a_n=\frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^n-\frac{1}{\sqrt{5}}{\left(\frac{1-\sqrt{5}}{2}\right)}^n$

Explanation:

Consider the given information.

Put the value of $n=1\text{to}7$

into given term:

  $\begin{array}{l}{a}_1=\frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^1-\frac{1}{\sqrt{5}}{\left(\frac{1-\sqrt{5}}{2}\right)}^1\\ a_1=\frac{1}{\sqrt{5}}\cdot \frac{1+\sqrt{5}-1+\sqrt{5}}{2}\\ a_1=\frac{1}{\sqrt{5}}\cdot \frac{2\sqrt{5}}{2}\\ a_1=1\end{array}$

  $\begin{array}{l}{a}_2=\frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^2-\frac{1}{\sqrt{5}}{\left(\frac{1-\sqrt{5}}{2}\right)}^2\\ a_2=\frac{1}{\sqrt{5}}\cdot \frac{1+5+2\sqrt{5}-1-5+2\sqrt{5}}{2}\\ a_2=\frac{1}{\sqrt{5}}\cdot \frac{4\sqrt{5}}{4}\\ a_2=1\end{array}$

  $\begin{array}{l}{a}_3=\frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^3-\frac{1}{\sqrt{5}}{\left(\frac{1-\sqrt{5}}{2}\right)}^3\\ a_3=\frac{1}{\sqrt{5}}\cdot \frac{1+3\sqrt{5}+15+5\sqrt{5}-1+3\sqrt{5}-15+5\sqrt{5}}{8}\\ a_3=\frac{1}{\sqrt{5}}\cdot \frac{16\sqrt{5}}{8}\\ a_3=2\end{array}$

  $\begin{array}{l}{a}_4=\frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^4-\frac{1}{\sqrt{5}}{\left(\frac{1-\sqrt{5}}{2}\right)}^4\\ a_4=\frac{1}{\sqrt{5}}\cdot \frac{1+4\sqrt{5}+30+20\sqrt{5}+25-1+4\sqrt{5}-30+20\sqrt{5}-25}{16}\\ a_4=\frac{1}{\sqrt{5}}\cdot \frac{16\sqrt{5}}{8}\\ a_4=3\\ a_5=\frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^5-\frac{1}{\sqrt{5}}{\left(\frac{1-\sqrt{5}}{2}\right)}^5\\ a_5=\frac{1}{\sqrt{5}}\cdot \frac{1+5\sqrt{5}+50+50\sqrt{5}+125-1+5\sqrt{5}-50+50\sqrt{5}-25\sqrt{5}}{32}\\ a_5=\frac{1}{\sqrt{5}}\cdot \frac{160\sqrt{5}}{32}\\ a_5=5\end{array}$

  $\begin{array}{l}{a}_6=\frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^6-\frac{1}{\sqrt{5}}{\left(\frac{1-\sqrt{5}}{2}\right)}^6\\ a_6=\frac{1}{\sqrt{5}}\cdot \frac{1+6\sqrt{5}+75+100\sqrt{5}+375+150\sqrt{5}+125-1+6\sqrt{5}-75+100\sqrt{5}-375+150\sqrt{5}-125}{64}\\ a_6=\frac{1}{\sqrt{5}}\cdot \frac{512\sqrt{5}}{64}\\ a_6=8\end{array}$

  $\begin{array}{l}{a}_7=\frac{1}{\sqrt{5}}{\left(\frac{1+\sqrt{5}}{2}\right)}^7-\frac{1}{\sqrt{5}}{\left(\frac{1-\sqrt{5}}{2}\right)}^7\\ a_7=\frac{1}{\sqrt{5}}\cdot \frac{\begin{array}{c}1+7\sqrt{5}+105+175\sqrt{5}+875+525\sqrt{5}+875+125\sqrt{5}-1+7\sqrt{5}-105+175\sqrt{5}-875+\\ 525\sqrt{5}-875+125\sqrt{5}\end{array}}{128}\\ a_7=\frac{1}{\sqrt{5}}\cdot \frac{1664\sqrt{5}}{128}\\ a_7=13\end{array}$

So, the term of the given sequence is the term of the Fibonacci sequence.