/*! 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 14 For the sequence \(a_{1}, a_{2},... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 14

For the sequence \(a_{1}, a_{2}, \ldots, a_{n}, \ldots,\) assume that \(a_{1}=1, a_{2}=1,\) and that for each \(n \in \mathbb{N}, a_{n+2}=\frac{1}{2}\left(a_{n+1}+\frac{2}{a_{n}}\right)\). (a) Calculate \(a_{3}\) through \(a_{6}\). (b) Prove that for each \(n \in \mathbb{N}, 1 \leq a_{n} \leq 2\).

Short Answer

Expert verified
We found that: \[a_3 = \frac{3}{2},\newline a_4 = \frac{7}{4},\newline a_5 = \frac{49}{24},\newline a_6 = \frac{393}{168}\] Furthermore, we proved that \(1 \leq a_n \leq 2\) for each \(n \in \mathbb{N}\) by using mathematical induction.

Step by step solution

01

Calculate \(a_{3}\)

Given \(a_{1} = 1\) and \(a_{2} = 1\), we have: \[a_{3} = \frac{1}{2}(a_2 + \frac{2}{a_1}) = \frac{1}{2}(1+\frac{2}{1}) = \frac{3}{2}\]
02

Calculate \(a_{4}\)

Using \(a_{2} = 1\) and \(a_{3} = \frac{3}{2}\), we have: \[a_{4} = \frac{1}{2}(a_3 + \frac{2}{a_2}) = \frac{1}{2}(\frac{3}{2} + \frac{2}{1}) = \frac{1}{2}(3.5) = \frac{7}{4}\]
03

Calculate \(a_{5}\)

Using \(a_{3} = \frac{3}{2}\) and \(a_{4} = \frac{7}{4}\), we have: \[a_{5} = \frac{1}{2}(a_4 + \frac{2}{a_3}) = \frac{1}{2}(\frac{7}{4}+\frac{2}{\frac{3}{2}}) = \frac{1}{2}(\frac{7}{4}+\frac{4}{3}) = \frac{49}{24}\]
04

Calculate \(a_{6}\)

Using \(a_{4} = \frac{7}{4}\) and \(a_{5} = \frac{49}{24}\), we have: \[a_{6} = \frac{1}{2}(a_5+\frac{2}{a_4}) = \frac{1}{2}(\frac{49}{24}+\frac{2}{\frac{7}{4}}) = \frac{1}{2}(\frac{49}{24}+\frac{8}{7}) = \frac{393}{168}\] So, we found the values for \(a_3, a_4, a_5,\) and \(a_6\) as: \[a_3 = \frac{3}{2},\newline a_4 = \frac{7}{4},\newline a_5 = \frac{49}{24},\newline a_6 = \frac{393}{168}\] #Part (b): Prove that \(1 \leq a_n \leq 2\) for each $n \in \mathbb{N}# We will use mathematical induction to prove the inequality.
05

Base cases: \(n = 1\) and \(n = 2\)

For \(n = 1\), we have \(a_1 = 1\), so \(1 \leq a_1 \leq 2\) holds. For \(n = 2\), we have \(a_2 = 1\), so \(1 \leq a_2 \leq 2\) also holds.
06

Inductive step

Suppose the inequality holds for all \(n \leq k\), i.e., \[1 \leq a_n \leq 2, \ \forall{n \in \mathbb{N}, \;n \leq k}\] Now, we need to show that \(1 \leq a_{k+1} \leq 2\). Using the recursive formula, we have: \[a_{k+1} = \frac{1}{2}\left(a_k+\frac{2}{a_{k-1}}\right)\] Since we know that \(1 \leq a_k \leq 2\) and \(1 \leq a_{k-1} \leq 2\), we can make the following observations: - \(a_k + \frac{2}{a_{k-1}} \geq 1 + \frac{2}{2} = 2\) - \(a_k + \frac{2}{a_{k-1}} \leq 2 + \frac{2}{1} = 4\) Now, using these observations, we have: \[\frac{1}{2}(a_k+\frac{2}{a_{k-1}}) \geq \frac{1}{2}(2) = 1\] \[\frac{1}{2}(a_k+\frac{2}{a_{k-1}}) \leq \frac{1}{2}(4) = 2\] Which implies that \(1 \leq a_{k+1} \leq 2\). Hence, the inequality holds for each \(n \in \mathbb{N}\) by the principle of mathematical induction.

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.

Mathematical Reasoning
Mathematical reasoning is the process by which we arrive at conclusions based on a structured set of principles or logic. It is a fundamental aspect of mathematics, as it underpins our ability to understand, justify, and solve problems in a rigorous way.

When faced with a mathematical statement or problem 鈥� such as proving the behavior of a sequence or its bounds 鈥� it's essential to use logical steps that others can follow and verify. In the context of sequences, mathematical reasoning involves understanding the relationship between consecutive terms and applying this relationship to derive properties of the sequence.

For example, if we are given that a sequence starts with certain values and follows a certain recursive pattern, we can use our reasoning skills to predict the behavior of that sequence without having to calculate every single term. This is powerful because it allows us to make broad claims about an infinite set of numbers with confidence, as seen in the provided exercise.

Good mathematical reasoning often requires a clear grasp of all part of the question, from comprehending initial conditions to recognizing the underlying structure or pattern. It also precedes the practice of proof writing, where one demonstrates the truthfulness of a conjecture or statement formally and logically.
Proof Writing
Proof writing is an essential skill in mathematics, especially when it comes to conveying and validating the correctness of mathematical statements. A strong proof serves as a rigorous argument that establishes the truth of a proposition or a theorem.

In the exercise provided, the task of proving that for each natural number the terms of the sequence fall between 1 and 2 requires the student to engage in proof writing. A common technique for demonstrating such properties of sequences is through the method of mathematical induction.

Mathematical Induction

Mathematical induction is a powerful proof technique that consists of two main parts: the base case and the inductive step. The base case checks the truth of the statement for the initial number(s), often starting with the smallest value in the domain of discourse (for example, natural numbers). After establishing the base case, the inductive step is used to show that if the statement holds for a given case, then it also holds for the next case.

This logical domino effect allows mathematicians to extend the truth from the base case through all elements of the set by inference. Therefore, learning to craft this type of argument not only builds a student's proof writing skills but also enhances their ability to think inductively about mathematical problems.
Recursive Sequences
Recursive sequences are sequences in which each term after the first one or few is generated from the preceding terms using a rule called a recurrence relation. These sequences are fascinating because they can generate complex and interesting patterns from simple rules.

In our exercise, the sequence is defined recursively via the relation \(a_{n+2} = \frac{1}{2}\left(a_{n+1}+\frac{2}{a_{n}}\right)\). Understanding and working with such sequences require recognizing the dependency of each term on its predecessors, a concept central to recursion.

Calculating Terms in Recursive Sequences

As demonstrated in the step by step solution, calculating successive terms of the sequence is straightforward once the recurrence is known. For recursive sequences, it's crucial to have the initial terms, as these are needed to kickstart the generation of subsequent terms. In this case, the sequence begins with \(a_{1} = 1\) and \(a_{2} = 1\), allowing the calculation of \(a_{3}\) through \(a_{6}\) as seen in the solution.

Moreover, the behavior of recursive sequences can often be predicted or bounded without calculating every term. This ability is especially important because it's typically impossible or impractical to write out all the terms of an infinite sequence. In the case of our exercise, proving a range between two constants for all terms introduces students to the concept that sequences can have their behavior characterized in the long term, which is a common theme in the study of sequences and series.

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

(a) For each natural number \(n, 1+4+7+\cdots+(3 n-2)=\frac{n(3 n-1)}{2}\). So we let \(P(n)\) be the open sentence $$ 1+4+7+\cdots+(3 n-2) $$ Using \(n=1,\) we see that \(3 n-2=1\) and hence, \(P(1)\) is true. We now assume that \(P(k)\) is true. That is, $$ 1+4+7+\cdots+(3 k-2)=\frac{k(3 k-1)}{2} $$ We then see that $$ \begin{aligned} 1+4+7+\cdots+(3 k-2)+(3(k+1)-2) &=\frac{(k+1)(3 k+2)}{2} \\ \frac{k(3 k-1)}{2}+(3 k+1) &=\frac{(k+1)(3 k+2)}{2} \\ \frac{\left(3 k^{2}-k\right)+(6 k+2)}{2} &=\frac{3 k^{2}+5 k+2}{2} \\ \frac{3 k^{2}+5 k+2}{2} &=\frac{3 k^{2}+5 k+2}{2} . \end{aligned} $$ We have thus proved that \(P(k+1)\) is true, and hence, we have proved the proposition. (b) For each natural number \(n, 1+4+7+\cdots+(3 n-2)=\frac{n(3 n-1)}{2}\). Proof. We will prove this proposition using mathematical induction. So we let $$ P(n)=1+4+7+\cdots+(3 n-2) $$Using \(n=1,\) we see that \(P(1)=1\) and hence, \(P(1)\) is true. We now assume that \(P(k)\) is true. That is, $$ 1+4+7+\cdots+(3 k-2)=\frac{k(3 k-1)}{2} $$ We then see that $$ \begin{aligned} P(k+1) &=1+4+7+\cdots+(3 k-2)+(3(k+1)-2) \\ &=\frac{k(3 k-1)}{2}+3(k+1)-2 \\ &=\frac{3 k^{2}-k+6 k+6-4}{2} \\ &=\frac{3 k^{2}+5 k+2}{2} \\ &=\frac{(k+1)(3 k+2)}{2} \end{aligned} $$ We have thus proved that \(P(k+1)\) is true, and hence, we have proved the proposition. (c) All dogs are the same breed. Proof. We will prove this proposition using mathematical induction. For each natural number \(n,\) we let \(P(n)\) be Any set of \(n\) dogs consists entirely of dogs of the same breed. We will prove that for each natural number \(n, P(n)\) is true, which will prove that all dogs are the same breed. A set with only one dog consists entirely of dogs of the same breed and, hence, \(P(1)\) is true. So we let \(k\) be a natural number and assume that \(P(k)\) is true, that is, that every set of \(k\) dogs consists of dogs of the same breed. Now consider a set \(D\) of \(k+1\) dogs, where $$ D=\left\\{d_{1}, d_{2}, \ldots, d_{k}, d_{k+1}\right\\} $$ If we remove the \(\operatorname{dog} d_{1}\) from the set \(D\), we then have a set \(D_{1}\) of \(k\) dogs, and using the assumption that \(P(k)\) is true, these dogs must all be of the same breed. Similarly, if we remove \(d_{k+1}\) from the set \(D,\) we again have a set \(D_{2}\) of \(k\) dogs, and these dogs must all be of the same breed. Since \(D=D_{1} \cup D_{2},\) we have proved that all of the dogs in \(D\) must be of the same breed. This proves that if \(P(k)\) is true, then \(P(k+1)\) is true and, hence, by mathematical induction, we have proved that for each natural number \(n,\) any set of \(n\) dogs consists entirely of dogs of the same breed.

Use mathematical induction to prove each of the following: (a) For each natural number \(n\) with \(n \geq 2,3^{n}>1+2^{n}\). (b) For each natural number \(n\) with \(n \geq 6,2^{n}>(n+1)^{2}\). (c) For each natural number \(n\) with \(n \geq 3,\left(1+\frac{1}{n}\right)^{n}

(a) Calculate the value of \(5^{n}-2^{n}\) for \(n=1, n=2, n=3, n=4, n=5\), and \(n=6\). (b) Based on your work in Part (a), make a conjecture about the values of \(5^{n}-2^{n}\) for each natural number \(n\). (c) Use mathematical induction to prove your conjecture in Part (b).

Compound Interest. Assume that \(R\) dollars is deposited in an account that has an interest rate of \(i\) for each compounding period. A compounding period is some specified time period such as a month or a year. For each integer \(n\) with \(n \geq 0,\) let \(V_{n}\) be the amount of money in an account at the end of the \(n\) th compounding period. Then $$ \begin{array}{rlrl} V_{1}=R+i \cdot R & V_{2} & =V_{1}+i \cdot V_{1} \\ =R(1+i) & & =(1+i) V_{1} \\ & = & (1+i)^{2} R . \end{array} $$ (a) Explain why \(V_{3}=V_{2}+i \cdot V_{2}\). Then use the formula for \(V_{2}\) to determine a formula for \(V_{3}\) in terms of \(i\) and \(R\). (b) Determine a recurrence relation for \(V_{n+1}\) in terms of \(i\) and \(V_{n}\). (c) Write the recurrence relation in Part ( \(22 \mathrm{~b}\) ) so that it is in the form of a recurrence relation for a geometric sequence. What is the initial term of the geometric sequence and what is the common ratio? (d) Use Proposition 4.14 to determine a formula for \(V_{n}\) in terms of \(I, R\), and \(n\).

Prove or disprove each of the following propositions: (a) For each \(n \in \mathbb{N}, \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}\). (b) For each natural number \(n\) with \(n \geq 3\), $$ \frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\cdots+\frac{1}{n(n+1)}=\frac{n-2}{3 n+3} $$ (c) For each \(n \in \mathbb{N}, 1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n(n+1)=\frac{n(n+1)(n+2)}{3}\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

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.