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Chapter 4: Problem 6

Prove Proposition 4.14. Let \(a, r \in \mathbb{R}\). If a geometric sequence is defined by \(a_{1}=a\) and for each \(n \in \mathbb{N}, a_{n+1}=r \cdot a_{n},\) then for each \(n \in \mathbb{N}\) \(a_{n}=a \cdot r^{n-1} .\)

Short Answer

Expert verified
In order to prove Proposition 4.14, we used induction. We first showed that the formula \(a_n = a \cdot r^{n-1}\) holds for the base case n = 1. Then, we assumed that the formula holds for an arbitrary positive integer "n" and demonstrated that it also holds for "n+1". Consequently, we have proven by induction that the formula \(a_n = a \cdot r^{n-1}\) holds for each \(n \in \mathbb{N}\) in the geometric sequence defined by \(a_{1}=a\) and for each \(n \in \mathbb{N}, a_{n+1}=r \cdot a_{n}\).

Step by step solution

01

Base case: n = 1

In this case, we have to show that the formula holds: \(a_n = a \cdot r^{n-1}\). For n = 1, we can write it as: \(a_1 = a \cdot r^{1-1}\). Now, \(r^{1-1} = r^0 = 1\), so \(a_1 = a \cdot 1 = a\). The initial term of the sequence is given as \(a_1 = a\) (by definition), which implies that our formula holds for the base case n = 1.
02

Inductive hypothesis

Now, we will assume that the formula holds for an arbitrary positive integer "n": \(a_n = a \cdot r^{n-1}\). This is the inductive hypothesis. Our goal is to show that it also holds for "n+1": \(a_{n+1} = a \cdot r^{(n+1)-1}\).
03

Inductive step

We know that \(a_{n+1} = r \cdot a_n\), according to the definition of the geometric sequence. We can substitute our inductive hypothesis (the formula for \(a_n\)) into this equation: \[a_{n+1} = r \cdot (a \cdot r^{n-1})\] Now, simplify the right-hand side by multiplying "r" through the parentheses: \[a_{n+1} = a \cdot r^{n}\] This equation is the same as our goal, which is to show \(a_{n+1} = a \cdot r^{(n+1)-1}\). Thus, if the formula holds for an arbitrary positive integer "n", it also holds for "n+1".
04

Conclusion

Since the formula holds for the base case n = 1, and the inductive step showed that if it holds for an arbitrary positive integer "n", it also holds for "n+1", we have proved by induction that Proposition 4.14 is true. The formula \(a_{n} = a \cdot r^{n-1}\) holds for each \(n \in \mathbb{N}\) in the geometric sequence defined by \(a_{1}=a\) and for each \(n \in \mathbb{N}, a_{n+1}=r \cdot a_{n}\).

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Key Concepts

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

Geometric Sequences
A geometric sequence is a series of numbers where each term is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio. This ratio remains constant throughout the sequence. For example, in the sequence 2, 6, 18, 54, ..., each term is three times the previous term, making the common ratio 3.

The general form of a geometric sequence can be expressed as follows:
  • The first term, denoted as \(a_1\), is arbitrary.
  • The nth term, \(a_n\), can be found by multiplying the first term \(a_1\) by the common ratio \(r\) to the power of \(n-1\): \(a_n = a_1 \times r^{n-1}\).
This defining property of geometric sequences allows for a straightforward determination of any term within the sequence given the first term and the common ratio.
Inductive Hypothesis
The inductive hypothesis is an assumption made in the process of mathematical induction, a powerful method for proving that a statement holds true for all natural numbers. When we use induction, we first show the base case (typically when \(n = 1\)) is true. Then we assume the statement is true for some arbitrary positive integer \(n\), which is called the inductive hypothesis.

For example, in the context of proving properties of geometric sequences, we might assume that the nth term is given by the formula \(a_n = a_1 \times r^{n-1}\) for some arbitrary \(n\). We use this assumption to show that if the formula hold for \(n\), it must also hold for \(n+1\). The inductive hypothesis serves as a bridge that allows us to extend the truth from one case to the next, forming a continuous chain of truth across all natural numbers.

Incorporating the inductive hypothesis into solution steps facilitates our understanding of how the sequence of operations extends the validity of our formula from one term to the next, cementing our grasp of the concept.
Mathematical Reasoning
Mathematical reasoning refers to the logical thought process involved in solving mathematical problems. It encompasses a range of techniques, from deduction to mathematical induction, which has been used in the earlier steps of the solution.

In our geometric sequence example, mathematical reasoning helps us understand how the proof works in a step-by-step fashion:
  • We begin with a base case to show that our statement is true in a simple instance.
  • We formulate an inductive hypothesis that assumes the statement is true for an arbitrary instance.
  • We then demonstrate that this assumption leads inevitably to the statement being true for the next instance (the inductive step).
  • Finally, we conclude that, because the statement holds for the base case and the induction step is valid, the statement must be true for all natural numbers.
This approach illustrates a fundamental aspect of mathematical reasoning: the ability to build upon established truths to reach a new understanding. Through mathematical induction, we prove propositions that appear to be intuitively true by linking them in an unbroken chain of logical deductions.

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Most popular questions from this chapter

Let \(x\) and \(y\) be integers. Prove that for each natural number \(n,(x-y)\) divides \(\left(x^{n}-y^{n}\right) .\) Explain why your conjecture in Exercise (11) is a special case of this result.

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 \(n \geq 3, L_{n}=f_{n+2}-f_{n-2}\).

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 \(n \in \mathbb{N}, 1 \leq a_{n} \leq 2\).

Use mathematical induction to prove each of the following: (a) For each natural number \(n, 3\) divides \(\left(4^{n}-1\right)\). (b) For each natural number \(n, 6\) divides \(\left(n^{3}-n\right)\).

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 \((1+2+3+\cdots+n) ?\)
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