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Chapter 4: Problem 13

For the sequence \(a_{1}, a_{2}, \ldots, a_{n}, \ldots,\) assume that \(a_{1}=1, a_{2}=3,\) and that for each \(n \in \mathbb{N}, a_{n+2}=3 a_{n+1}-2 a_{n}\) (a) Calculate \(a_{3}\) through \(a_{6}\). (b) Make a conjecture for a formula for \(a_{n}\) for each \(n \in \mathbb{N}\). (c) Prove that your conjecture in Exercise (13b) is correct.

Short Answer

Expert verified
The sequence is \(1, 3, 7, 15, 31, 63, \ldots\), and the formula for the terms in the sequence is \(a_n=2^n-1\) for each \(n \in \mathbb{N}\). This conjecture is proved using mathematical induction.

Step by step solution

01

Calculate \(a_{3}\) through \(a_{6}\) using the recursion formula

We have the recursion formula \(a_{n+2}=3 a_{n+1}-2 a_{n}\). Let's use it to find \(a_{3}\) through \(a_{6}\): For \(a_{3}\): \(a_{3}=3a_{2}-2a_{1}=3(3)-2(1)=9-2=7\) For \(a_{4}\): \(a_{4}=3a_{3}-2a_{2}=3(7)-2(3)=21-6=15\) For \(a_{5}\): \(a_{5}=3a_{4}-2a_{3}=3(15)-2(7)=45-14=31\) For \(a_{6}\): \(a_{6}=3a_{5}-2a_{4}=3(31)-2(15)=93-30=63\) So we have the sequence: \(1, 3, 7, 15, 31, 63, \ldots\)
02

Make a conjecture for a formula for \(a_{n}\)

Looking at the sequence, it seems to be each term is one less than the next power of 2: \(a_1=2^1-1=1; a_2=2^2-1=3; a_3=2^3-1=7;\) and so on. This suggests the following conjecture: \(a_n=2^n-1\) for each \(n \in \mathbb{N}\).
03

Prove the conjecture using mathematical induction

We will prove our conjecture \(a_n=2^n-1\) using mathematical induction. Base case (n = 1): We have \(a_1=1\) and our conjecture says \(a_1=2^1-1=1\), which is true. Inductive step: Assume the formula holds true for some \(n=k\), which means \(a_k=2^k-1\). Our goal is to show that the formula also holds for \(n=k+1\), which means we need to show that \(a_{k+1}=2^{k+1}-1\). We are given the recursion formula \(a_{n+2}=3a_{n+1}-2a_{n}\). Let's apply it to find \(a_{k+1}\): \(a_{k+1}=3a_k-2a_{k-1}\) By the inductive assumption, we know \(a_k=2^k-1\) and \(a_{k-1}=2^{k-1}-1\). Plugging these into the equation, we get: \(a_{k+1}=3(2^k-1)-2(2^{k-1}-1)\) Now, we can simplify this expression: \(a_{k+1}=3\cdot2^k-3-2^k+2=2^{k+1}-1\) So, the formula \(a_n=2^n-1\) holds for \(n=k+1\) as well. Hence, by mathematical induction, our conjecture is proved for all \(n \in \mathbb{N}\).

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Key Concepts

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

Sequence
A sequence is a set of numbers ordered in a specific manner, often following a recognisable pattern or rule. In the problem, the sequence begins with 1 and 3, forming the basis from which other terms are generated. This setup is known as a starting point in sequences.
In mathematical terms, sequences can be defined either explicitly or recursively. In an explicit sequence, each term is defined independently of the others, often involving a formula based on the position of the term. However, in a recursive sequence like ours, each term after the initial ones is determined by a relationship that uses previous terms.
This specific problem uses just such a relationship to define a sequence, where each term is calculated using the previous two terms. Understanding the initial conditions is crucial as they help establish the pattern that the sequence will follow.
Recursion Formula
A recursion formula is a procedure or rule that generates each term of a sequence using previous terms. In the given problem, the sequence is governed by the recursion formula: \(a_{n+2} = 3a_{n+1} - 2a_n\). This shows that every term from the third onwards can be calculated with this formula.
Utilizing recursion simplifies calculating the progression of numbers without needing a direct formula for \(a_n\). To apply this, you start with known terms (\(a_1 = 1\), \(a_2 = 3\)) and iterate, successively replacing previous values to compute later terms like \(a_3\), \(a_4\), and so on.
It's essential in recursive formulas to correctly use the previously derived terms, as any mistake might lead to a breakdown of the sequence. It鈥檚 a powerful tool in sequences that exhibit predictable patterns based on previous values.
Conjecture
A conjecture in mathematics is an educated guess or hypothesis about what a pattern might be, based on observing specific cases. Due to the nature of sequences, especially recursive ones, it's often possible to discern a regular pattern.
In the exercise, after finding the initial terms, a pattern emerges: each term appears to be one less than a power of two. This observation leads to the conjecture that \(a_n = 2^n - 1\) for the sequence.
Conjecturing is a crucial step in advancing mathematical problems as it proposes a general rule that can potentially describe the pattern. The validity of this conjecture is then tested by proving it, such as through mathematical induction. This way, the pattern is verified for all applicable numbers, ensuring that the conjecture is not only an educated guess but a valid mathematical fact for the entirety of the sequence.

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Most popular questions from this chapter

The quadratic formula can be used to show that \(\alpha=\frac{1+\sqrt{5}}{2}\) and \(\beta=\frac{1-\sqrt{5}}{2}\) are the two real number solutions of the quadratic equation \(x^{2}-x-1=0\). Notice that this implies that $$ \begin{array}{l} \alpha^{2}=\alpha+1, \text { and } \\ \beta^{2}=\beta+1 \end{array} $$ It may be surprising to find out that these two irrational numbers are closely related to the Fibonacci numbers. (a) Verify that \(f_{1}=\frac{\alpha^{1}-\beta^{1}}{\alpha-\beta}\) and that \(f_{2}=\frac{\alpha^{2}-\beta^{2}}{\alpha-\beta}\). (b) (This part is optional, but it may help with the induction proof in part (c).) Work with the relation \(f_{3}=f_{2}+f_{1}\) and substitute the expressions for \(f_{1}\) and \(f_{2}\) from part (a). Rewrite the expression as a single fraction and then in the numerator use \(\alpha^{2}+\alpha=\alpha(\alpha+1)\) and a similar equation involving \(\beta .\) Now prove that \(f_{3}=\frac{\alpha^{3}-\beta^{3}}{\alpha-\beta}\).(c) Use induction to prove that for each natural number \(n,\) if \(\alpha=\frac{1+\sqrt{5}}{2}\) and \(\beta=\frac{1-\sqrt{5}}{2},\) then \(f_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta} .\) Note: This formula for the \(n^{t h}\) Fibonacci number is known as Binet's formula, named after the French mathematician Jacques Binet ( \(1786-1856\) ).

Prove that for each natural number \(n\), any set with \(n\) elements has \(\frac{n(n-1)}{2}\) two-element subsets.

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 \(n\).

In calculus, it can be shown that $$ \begin{array}{l} \int \sin ^{2} x d x=\frac{x}{2}-\frac{1}{2} \sin x \cos x+c \quad \text { and } \\ \int \cos ^{2} x d x=\frac{x}{2}+\frac{1}{2} \sin x \cos x+c \end{array} $$ Using integration by parts, it is also possible to prove that for each natural number \(n,\) $$ \begin{aligned} \int \sin ^{n} x d x &=-\frac{1}{n} \sin ^{n-1} x \cos x+\frac{n-1}{n} \int \sin ^{n-2} x d x \text { and } \\ \int \cos ^{n} x d x &=\frac{1}{n} \cos ^{n-1} x \sin x+\frac{n-1}{n} \int \cos ^{n-2} x d x \end{aligned} $$ (a) Determine the values of $$ \int_{0}^{\pi / 2} \sin ^{2} x d x $$ and $$ \int_{0}^{\pi / 2} \sin ^{4} x d x $$ (b) Use mathematical induction to prove that for each natural number \(n\), $$ \begin{aligned} \int_{0}^{\pi / 2} \sin ^{2 n} x d x &=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots(2 n)} \frac{\pi}{2} \text { and } \\ \int_{0}^{\pi / 2} \sin ^{2 n+1} x d x &=\frac{2 \cdot 4 \cdot 6 \cdots(2 n)}{1 \cdot 3 \cdot 5 \cdots(2 n+1)} \end{aligned} $$ These are known as the Wallis sine formulas. (c) Use mathematical induction to prove that $$ \begin{aligned} \int_{0}^{\pi / 2} \cos ^{2 n} x d x &=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots(2 n)} \frac{\pi}{2} \quad \text { and } \\ \int_{0}^{\pi / 2} \cos ^{2 n+1} x d x &=\frac{2 \cdot 4 \cdot 6 \cdots(2 n)}{1 \cdot 3 \cdot 5 \cdots(2 n+1)} \end{aligned} $$ These are known as the Wallis cosine formulas.

The Future Value of an Ordinary Annuity. For an ordinary annuity, \(R\) dollars is deposited in an account at the end of each compounding period. It is assumed that the interest rate, \(i,\) per compounding period for the account remains constant. Let \(S_{t}\) represent the amount in the account at the end of the \(t\) th compounding period. \(S_{t}\) is frequently called the future value of the ordinary annuity. So \(S_{1}=R\). To determine the amount after two months, we first note that the amount after one month will gain interest and grow to \((1+i) S_{1} .\) In addition, a new deposit of \(R\) dollars will be made at the end of the second month. So $$ S_{2}=R+(1+i) S_{1} $$ (a) For each \(n \in \mathbb{N},\) use a similar argument to determine a recurrence relation for \(S_{n+1}\) in terms of \(R, i,\) and \(S_{n}\). (b) By recognizing this as a recursion formula for a geometric series, use Proposition 4.16 to determine a formula for \(S_{n}\) in terms of \(R, i,\) and \(n\) that does not use a summation. Then show that this formula can be written as $$ S_{n}=R\left(\frac{(1+i)^{n}-1}{i}\right) $$ (c) What is the future value of an ordinary annuity in 20 years if \(\$ 200\) dollars is deposited in an account at the end of each month where the interest rate for the account is \(6 \%\) per year compounded monthly? What is the amount of interest that has accumulated in this account during the 20 years?
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