/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 The Lucas numbers are a sequence... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 18

The Lucas numbers are a sequence of natural numbers \(L_{1}, L_{2}, L_{3}, \ldots, L_{n}, \ldots\) which are defined recursively as follows: \- \(L_{1}=1\) and \(L_{2}=3,\) and \- For each natural number \(n, L_{n+2}=L_{n+1}+L_{n}\). List the first 10 Lucas numbers and the first ten Fibonacci numbers and then prove each of the following propositions. The Second Principle of Mathematical Induction may be needed to prove some of these propositions. (a) For each natural number \(n, L_{n}=2 f_{n+1}-f_{n}\) (b) For each \(n \in \mathbb{N}\) with \(n \geq 2,5 f_{n}=L_{n-1}+L_{n+1}\). (c) For each \(n \in \mathbb{N}\) with \(n \geq 3, L_{n}=f_{n+2}-f_{n-2}\).

Short Answer

Expert verified
In conclusion, we have found the first ten Lucas numbers and the first ten Fibonacci numbers. We also proved the following propositions: (a) For each natural number \(n\), \(L_n = 2F_{n+1} - F_n\). (b) For each \(n \in \mathbb{N}\) with \(n \geq 2\), \(5F_n = L_{n-1} + L_{n+1}\). (c) For each \(n \in \mathbb{N}\) with \(n \geq 3\), \(L_n = F_{n+2} - F_{n-2}\). We used the mathematical induction method to prove all the three propositions successfully.

Step by step solution

01

First ten Lucas numbers

We are given \(L_1 = 1\) and \(L_2 = 3\). We can use the recursive definition \(L_{n+2} = L_{n+1} + L_n\) to calculate the first 10 Lucas numbers: 1. \(L_1 = 1\) 2. \(L_2 = 3\) 3. \(L_3 = L_1 + L_2 = 4\) 4. \(L_4 = L_2 + L_3 = 7\) 5. \(L_5 = L_3 + L_4 = 11\) 6. \(L_6 = L_4 + L_5 = 18\) 7. \(L_7 = L_5 + L_6 = 29\) 8. \(L_8 = L_6 + L_7 = 47\) 9. \(L_9 = L_7 + L_8 = 76\) 10. \(L_{10} = L_8 + L_9 = 123\)
02

First ten Fibonacci numbers

We are given \(F_1 = 1\), \(F_2 = 1\), and we can use the recursive definition \(F_{n+2} = F_{n+1} + F_n\): 1. \(F_1 = 1\) 2. \(F_2 = 1\) 3. \(F_3 = F_1 + F_2 = 2\) 4. \(F_4 = F_2 + F_3 = 3\) 5. \(F_5 = F_3 + F_4 = 5\) 6. \(F_6 = F_4 + F_5 = 8\) 7. \(F_7 = F_5 + F_6 = 13\) 8. \(F_8 = F_6 + F_7 = 21\) 9. \(F_9 = F_7 + F_8 = 34\) 10. \(F_{10} = F_8 + F_9 = 55\) ##Proof for Proposition (a)## We will prove that for each natural number \(n\), \(L_n = 2F_{n+1} - F_n\) using mathematical induction.
03

Base Case (a)

Check for \(n = 1\): \(L_1 = 1\), and \(2F_{1+1} - F_1 = 2F_2 - F_1 = 2(1) - 1 = 1\). It's true for \(n = 1\). Now check for \(n = 2\): \(L_2 = 3\), and \(2F_{2+1} - F_2 = 2F_3 - F_2 = 2(2) - 1 = 3\). It's true for \(n = 2\).
04

Inductive Step (a)

Assume that the proposition is true for \(n = k\) and \(n=k+1\). That is, assume: 1. \(L_k = 2F_{k+1} - F_k\), 2. \(L_{k+1} = 2F_{k+2} - F_{k+1}\). We want to show that the proposition is also true for \(n = k+2\), i.e., \(L_{k+2} = 2F_{k+3} - F_{k+2}\). By the recursive definition of Lucas numbers: \(L_{k+2} = L_{k+1} + L_k\). By our assumption: \(L_{k+2} = (2F_{k+2} - F_{k+1}) + (2F_{k+1} - F_k) = 2(F_{k+2} + F_{k+1}) - (F_{k+1} + F_k)\). Since \(F_{k+3} = F_{k+2} + F_{k+1}\) by the recursive definition of Fibonacci numbers, we can rewrite the equation as: \(L_{k+2} = 2F_{k+3} - F_{k+2}\). This completes the induction proof for Proposition (a). ##Proof for Proposition (b)## We will show that for each \(n \in \mathbb{N}\) with \(n \geq 2, 5F_n = L_{n-1} + L_{n+1}\), using mathematical induction.
05

Base Case (b)

Check for \(n=2\): \(5F_2 = 5(1) = 5\) \(L_{2-1} + L_{2+1} = L_1 + L_3 = 1 + 4 = 5\).
06

Inductive Step (b)

Assume that the proposition is true for \(n=k\): \(5F_k = L_{k-1} + L_{k+1}\). We want to show that it's also true for \(n=k+1\). Rearrange the assumption equation: $$L_{k+1} = 5F_k - L_{k-1} \surd (1).$$ By the recursive definition of Lucas numbers: $$L_{k+2} = L_{k+1} + L_k \surd (2).$$ By adding \((1) + (2)\), we get: $$L_{k+2} + L_{k-1} = (5F_k - L_{k-1}) + L_k + (5F_{k+1} - L_k) = 5(F_k + F_{k+1}).$$ Since \(F_{k+2} = F_{k+1} + F_k\): $$L_{k+2} + L_{k-1} = 5F_{k+2}.$$ Thus, the proposition is also true for \(n=k+1\). Therefore, the induction proof for Proposition (b) is complete. ##Proof for Proposition (c)## We will show that for each \(n \in \mathbb{N}\) with \(n \geq 3\), \(L_n = F_{n+2} - F_{n-2}\), using mathematical induction.
07

Base Case (c)

Check for \(n=3\): \(L_3 = 4\) \(F_{3+2} - F_{3-2} = F_5 - F_1 = 5 - 1 = 4\).
08

Inductive Step (c)

Assume that the proposition is true for \(n=k\): \(L_k = F_{k+2} - F_{k-2}\). We want to show that it's also true for \(n=k+1\). Rearrange the assumption equation: $$F_{k+2} = L_k + F_{k-2} \surd (1).$$ Substitute the result of Proposition (a) to the recursive definition of Lucas numbers: $$L_{k+1} = L_k + L_{k-1} = (2F_{k+1} - F_{k}) + (2F_{k} - F_{k-2}) = 2(F_{k} + F_{k+1}) - F_{k-2}.$$ Since \(F_{k+2} = F_{k+1} + F_{k}\), $$L_{k+1} = F_{k+3} - F_{k-1}.$$ Thus, the proposition is also true for \(n=k+1\). Therefore, the induction proof for Proposition (c) is complete.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lucas Numbers
Lucas Numbers, similar to the better-known Fibonacci sequence, form an intriguing series of integers that have been studied for their beautiful mathematical properties. Defined recursively, the first two Lucas numbers are set to specific values (\(L_1 = 1\) and \(L_2 = 3\)), and each subsequent number is the sum of the two preceding numbers (\(L_{n+2} = L_{n+1} + L_{n}\)).

While these numbers might seem arbitrary, they indeed exhibit a range of unique relationships with the Fibonacci numbers. For instance, some properties highlighting the connections between the two sequences, as outlined in the original exercise, can be proven using mathematical induction. By listing the first ten terms, we can get a visual impression of the growth of the Lucas numbers, which follow an increasing pattern similar to Fibonacci's but start from different initial conditions.
Fibonacci Sequence
The Fibonacci Sequence is a well-known series of numbers in which each number after the initial two is the sum of the two preceding ones. Typically starting with \(F_1 = 1\) and \(F_2 = 1\), the sequence continues with \(F_3 = 2\), \(F_4 = 3\), \(F_5 = 5\), and so on, following the rule \(F_{n+2} = F_{n+1} + F_n\).

It's notable for appearing in various natural phenomena, from the spirals of shells to the branching of trees to even the reproductive patterns of bees! The sequence illustrates a fundamental principle of growth patterns in nature and serves as a classic example of a recursive sequence in mathematics, being both simple to define and rich in applications.
Recursive Sequences
Recursive sequences are fundamental constructs in mathematics where each term is defined as a function of its preceding terms. Both Lucas and Fibonacci sequences serve as quintessential examples. Recursive definitions are not only crucial for generating sequences but are also a powerful tool in computer science, particularly in the design of algorithms.

Understanding recursive sequences is essential to grasp other mathematical concepts, such as series, fractals, and even certain aspects of mathematical proofs. They illustrate the concept of 'self-similarity', where a pattern repeats itself at different scales, and they form an integral part of the field of discrete mathematics.
Proof Techniques
Proof techniques are essential tools in mathematics that allow us to validate the truth of various statements. Mathematical induction, in particular, is a proof technique often used to prove statements about integers, especially when dealing with recursive sequences. Induction involves proving a base case and then showing that if the statement holds for an arbitrary case \(n = k\), it also holds for the next case \(n = k+1\).

Performing an inductive step involves critical reasoning and manipulation of algebraic expressions to achieve a form that matches the original assumption, thus confirming the property for all numbers in the sequence. Inductive proofs are powerful because they effectively demonstrate infinite cases of a proposition with a finite and logical process.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Instead of using induction, we can sometimes use previously proven results about a summation to obtain results about a different summation. (a) Use the result in Progress Check 4.3 to prove the following proposition: For each natural number \(n, 3+6+9+\cdots+3 n=\frac{3 n(n+1)}{2}\) (b) Subtract \(n\) from each side of the equation in Part (a). On the left side of this equation, explain why this can be done by subtracting 1 from each term in the summation. (c) Algebraically simplify the right side of the equation in Part (b) to obtain a formula for the sum \(2+5+8+\cdots+(3 n-1)\). Compare this to Exercise (3a).

Is the following proposition true or false? Justify your conclusion. For each natural number \(n,\left(\frac{n^{3}}{3}+\frac{n^{2}}{2}+\frac{7 n}{6}\right)\) is a natural number.

Based on the results in Progress Check 4.3 and Exercise \((3 \mathrm{c})\), if \(n \in \mathbb{N},\) is there any conclusion that can be made about the relationship between the \(\operatorname{sum}\left(1^{3}+2^{3}+3^{3}+\cdots+n^{3}\right)\) and the sum \((1+2+3+\cdots+n) ?\)

One of the most interesting results in trigonometry is De Moivre's Theorem, which relates the complex number \(i\) to the trigonometric functions. Recall that the number \(i\) is a complex number whose square is \(-1,\) that is, \(i^{2}=-1 .\) One version of the theorem can be stated as follows: If \(x\) is a real number, then for each nonnegative integer \(n,\) $$ [\cos x+i(\sin x)]^{n}=\cos (n x)+i(\sin (n x)) $$ This theorem is named after Abraham de Moivre \((1667-1754),\) a French mathematician. (a) The proof of De Moivre's Theorem requires the use of the trigonometric identities for the sine and cosine of the sum of two angles. Use the Internet or a book to find identities for \(\sin (\alpha+\beta)\) and \(\cos (\alpha+\beta)\). (b) To get a sense of how things work, expand \([\cos x+i(\sin x)]^{2}\) and write the result in the form \(a+b i\). Then use the identities from part (1) to prove that \([\cos x+i(\sin x)]^{2}=\cos (2 x)+i(\sin (2 x))\) (c) Use mathematical induction to prove De Moivre's Theorem.

(a) Why is it not possible to use mathematical induction to prove a proposition of the form $$ (\forall x \in \mathbb{Q})(P(x)) $$ where \(P(x)\) is some predicate? (b) Why is it not possible to use mathematical induction to prove a proposition of the form For each real number \(x\) with \(x \geq 1, P(x)\), where \(P(x)\) is some predicate?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.