McDougal Littell Jurgensen Geometry: Student Edition Geometry
5th Edition
ISBN: 9780395977279
Author: Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Publisher: Houghton Mifflin Company College Division
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Question
Chapter 7.4, Problem 1AE
To determine
To calculate: The value of ratios of two continuous terms in a fibonacci sequence
Expert Solution & Answer
Answer to Problem 1AE
The value of ratios of two continuous terms in a fibonacci sequence
Explanation of Solution
Given information:
The fibonacci number sequence is given as
There is one computer programme is also given to run it.
Formula used:
In the fibonacci number sequence the next term is sum of the last two terms, which can be represents by this,
Calculation:
Consider the given computer program and run it in the operating system.
Now, result comes out to be,
| Number | n-th Fibonacci Number | f(n)/f(n-1) | f(n-1)/f(n) | f(n)/f(n-1) | Phi=1.618… | f(n-1)/f(n) | Phi=0.618… | |
| f-0 | 1 | |||||||
| f-1 | 1 | 1 | 1 | 1 | 1.618034 | 1 | 0.618034 | |
| f-2 | 2 | 2 | 0.5 | 2 | 1.618034 | 0.5 | 0.618034 | |
| f-3 | 3 | 1.5 | 0.666667 | 1.5 | 1.618034 | 0.666667 | 0.618034 | |
| f-4 | 5 | 1.666667 | 0.6 | 1.666667 | 1.618034 | 0.6 | 0.618034 | |
| f-5 | 8 | 1.6 | 0.625 | 1.6 | 1.618034 | 0.625 | 0.618034 | |
| f-6 | 13 | 1.625 | 0.615385 | 1.625 | 1.618034 | 0.615385 | 0.618034 | |
| f-7 | 21 | 1.615385 | 0.619048 | 1.615385 | 1.618034 | 0.619048 | 0.618034 | |
| f-8 | 34 | 1.619048 | 0.617647 | 1.619048 | 1.618034 | 0.617647 | 0.618034 | |
| f-9 | 55 | 1.617647 | 0.618182 | 1.617647 | 1.618034 | 0.618182 | 0.618034 | |
| f-10 | 89 | 1.618182 | 0.617978 | 1.618182 | 1.618034 | 0.617978 | 0.618034 | |
| f-11 | 144 | 1.617978 | 0.618056 | 1.617978 | 1.618034 | 0.618056 | 0.618034 | |
| f-12 | 233 | 1.618056 | 0.618026 | 1.618056 | 1.618034 | 0.618026 | 0.618034 | |
| f-13 | 377 | 1.618026 | 0.618037 | 1.618026 | 1.618034 | 0.618037 | 0.618034 | |
| f-14 | 610 | 1.618037 | 0.618033 | 1.618037 | 1.618034 | 0.618033 | 0.618034 | |
| f-15 | 987 | 1.618033 | 0.618034 | 1.618033 | 1.618034 | 0.618034 | 0.618034 | |
| f-16 | 1597 | 1.618034 | 0.618034 | 1.618034 | 1.618034 | 0.618034 | 0.618034 | |
| f-17 | 2584 | 1.618034 | 0.618034 | 1.618034 | 1.618034 | 0.618034 | 0.618034 | |
| f-18 | 4181 | 1.618034 | 0.618034 | 1.618034 | 1.618034 | 0.618034 | 0.618034 |
It is very clear from the above table that ratio between two continuous successor and Predecessor term is approximate equal to 1.618.
Therefore, the value of ratios of two continuous terms in a fibonacci sequence
Chapter 7 Solutions
McDougal Littell Jurgensen Geometry: Student Edition Geometry
Ch. 7.1 - Prob. 1CECh. 7.1 - Prob. 2CECh. 7.1 - Prob. 3CECh. 7.1 - Prob. 4CECh. 7.1 - Prob. 5CECh. 7.1 - Prob. 6CECh. 7.1 - Prob. 7CECh. 7.1 - Prob. 8CECh. 7.1 - Prob. 9CECh. 7.1 - Prob. 10CE
Ch. 7.1 - Prob. 11CECh. 7.1 - Prob. 12CECh. 7.1 - Prob. 13CECh. 7.1 - Prob. 14CECh. 7.1 - Prob. 15CECh. 7.1 - Prob. 16CECh. 7.1 - Prob. 17CECh. 7.1 - Prob. 18CECh. 7.1 - Prob. 1WECh. 7.1 - Prob. 2WECh. 7.1 - Prob. 3WECh. 7.1 - Prob. 4WECh. 7.1 - Prob. 5WECh. 7.1 - Prob. 6WECh. 7.1 - Prob. 7WECh. 7.1 - Prob. 8WECh. 7.1 - Prob. 9WECh. 7.1 - Prob. 10WECh. 7.1 - Prob. 11WECh. 7.1 - Prob. 12WECh. 7.1 - Prob. 13WECh. 7.1 - Prob. 14WECh. 7.1 - Prob. 15WECh. 7.1 - Prob. 16WECh. 7.1 - Prob. 17WECh. 7.1 - Prob. 18WECh. 7.1 - Prob. 19WECh. 7.1 - Prob. 20WECh. 7.1 - Prob. 21WECh. 7.1 - Prob. 22WECh. 7.1 - Prob. 23WECh. 7.1 - Prob. 24WECh. 7.1 - Prob. 25WECh. 7.1 - Prob. 26WECh. 7.1 - Prob. 27WECh. 7.1 - Prob. 28WECh. 7.1 - Prob. 29WECh. 7.1 - Prob. 30WECh. 7.1 - Prob. 31WECh. 7.1 - Prob. 32WECh. 7.1 - Prob. 33WECh. 7.1 - Prob. 34WECh. 7.1 - Prob. 35WECh. 7.1 - Prob. 36WECh. 7.2 - Prob. 1CECh. 7.2 - Prob. 2CECh. 7.2 - Prob. 3CECh. 7.2 - Prob. 4CECh. 7.2 - Prob. 5CECh. 7.2 - Prob. 6CECh. 7.2 - Prob. 7CECh. 7.2 - 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Prob. 43WECh. 7.3 - Prob. 1CECh. 7.3 - Prob. 2CECh. 7.3 - Prob. 3CECh. 7.3 - Prob. 4CECh. 7.3 - Prob. 5CECh. 7.3 - Prob. 6CECh. 7.3 - Prob. 7CECh. 7.3 - Prob. 8CECh. 7.3 - Prob. 9CECh. 7.3 - Prob. 10CECh. 7.3 - Prob. 1WECh. 7.3 - Prob. 2WECh. 7.3 - Prob. 3WECh. 7.3 - Prob. 4WECh. 7.3 - Prob. 5WECh. 7.3 - Prob. 6WECh. 7.3 - Prob. 7WECh. 7.3 - Prob. 8WECh. 7.3 - Prob. 9WECh. 7.3 - Prob. 10WECh. 7.3 - Prob. 11WECh. 7.3 - Prob. 12WECh. 7.3 - Prob. 13WECh. 7.3 - Prob. 14WECh. 7.3 - Prob. 15WECh. 7.3 - Prob. 16WECh. 7.3 - Prob. 17WECh. 7.3 - Prob. 18WECh. 7.3 - Prob. 19WECh. 7.3 - Prob. 20WECh. 7.3 - Prob. 21WECh. 7.3 - Prob. 22WECh. 7.3 - Prob. 23WECh. 7.3 - Prob. 24WECh. 7.3 - Prob. 25WECh. 7.3 - Prob. 26WECh. 7.3 - Prob. 27WECh. 7.3 - Prob. 28WECh. 7.3 - Prob. 29WECh. 7.3 - Prob. 30WECh. 7.3 - Prob. 31WECh. 7.3 - Prob. 32WECh. 7.3 - Prob. 33WECh. 7.3 - Prob. 34WECh. 7.3 - Prob. 35WECh. 7.3 - Prob. 36WECh. 7.3 - Prob. 37WECh. 7.3 - Prob. 1ST1Ch. 7.3 - Prob. 2ST1Ch. 7.3 - Prob. 3ST1Ch. 7.3 - 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