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Chapter 2: Problem 6

Consider the three sequences below. For each, find a recursive definition. How are these sequences related? (a) \(2,4,6,10,16,26,42, \ldots .\) (b) \(5,6,11,17,28,45,73, \ldots\) (c) \(0,0,0,0,0,0,0, \ldots\)

Short Answer

Expert verified
Sequence (a): \( a(1) = 2, a(n) = a(n-1) + \sum_{i=1}^{n-2}a(i) + 2 \text{ for } n > 2. \) Sequence (b): \( b(1) = 5, b(n) = b(n-1) + \sum_{i=1}^{n-2}b(i) + 3 \text{ for } n > 2. \) Sequence (c): \( c(n) = 0 \text{ for all } n. \) Sequences (a) and (b) are each defined by the sum of all preceding terms plus a constant, whereas sequence (c) is simply a constant zero sequence.

Step by step solution

01

Analyzing Sequence (a)

First, let's find the pattern in sequence (a). Sequence (a) starts at 2, and each subsequent term after the second is the sum of all previous terms plus 2. This means to find the nth term for n > 2, we add the (n - 1)th term to the sum of all previous terms, starting from the first term, and then add 2.
02

Writing the Recursive Formula for Sequence (a)

Let's denote the nth term of sequence (a) as 'a(n)'. The recursive definition for sequence (a) will be:\[ a(1) = 2, \; a(n) = a(n-1) + \sum_{i=1}^{n-2}a(i) + 2 \text{ for } n > 2. \]
03

Analyzing Sequence (b)

Similar to sequence (a), sequence (b) also starts with a first term and each subsequent term is the sum of all previous terms plus 3. To find the nth term for n > 1, we add the (n - 1)th term to the sum of all earlier terms, then add 3.
04

Writing the Recursive Formula for Sequence (b)

Let 'b(n)' represent the nth term of sequence (b). The recursive definition would be:\[ b(1) = 5, \; b(n) = b(n-1) + \sum_{i=1}^{n-2}b(i) + 3 \text{ for } n > 2. \]
05

Analyzing Sequence (c)

Sequence (c) is composed completely of zeros. This means that each term is the same as its preceding term, which is zero.
06

Writing the Recursive Formula for Sequence (c)

The nth term 'c(n)' of sequence (c) can be defined recursively by: \[ c(n) = 0 \text{ for all } n. \]
07

Finding the Relationship Among the Sequences

By examining the recursive definitions, sequences (a) and (b) are similar in that each term is formed by the sum of all previous terms plus a constant (2 for sequence (a) and 3 for sequence (b)), while sequence (c) is essentially a zero sequence with no changes.

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Key Concepts

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

Sequence Patterns
Sequence patterns are the underlying rules that determine the progression of elements in a mathematical sequence. These patterns can be quite simple, such as a repeating element, or complex, involving arithmetic or geometric operations applied to previous terms. Understanding a sequence's pattern is crucial for determining the relationship between its terms and for predicting subsequent elements in the sequence.

For instance, in sequence (a) from the exercise (\(2,4,6,10,16,26,42, \rightarrow \)), the pattern involves summing all the previous terms and then adding 2 to get the next term. In general, recognizing sequence patterns requires careful observation and sometimes necessitates the calculation of differences between terms, or the application of other mathematical strategies to uncover the rule that governs the sequence's growth.
Recursive Formula
A recursive formula expresses each term of a sequence in relation to the terms before it. For a given sequence, the recursive formula provides a method to compute the nth term based on its predecessors rather than having a direct formula that can calculate the term independently.

In the textbook example, sequence (a) and (b) are defined through a recursive process. For sequence (a), the formula is given by \[ a(1) = 2, \: a(n) = a(n-1) + \sum_{i=1}^{n-2}a(i) + 2 \: \text{for} \: n > 2. \.\] This equation translates to starting the sequence with 2 and then, to find any term in the sequence beyond the second term, one must add the previous term, the sum of all preceding terms, and then add 2. This recursive nature allows for building the sequence one term at a time, based on the previously computed terms.
Mathematical Induction
Mathematical induction is a method of proof used in mathematics to establish the truth of an infinite number of cases. It is particularly useful for proving properties of recursively defined sequences and other mathematical objects defined by recurrence relations.

Induction involves two main steps: the base case and the inductive step. The base case verifies that the statement is true for the initial term of the sequence. Following this, the inductive step demonstrates that if the statement holds for an arbitrary term of the sequence, it must also hold for the next term.

In the context of the textbook exercise, mathematical induction could be employed to prove that the recursive formulas for sequences (a) and (b) provide correct terms for all positive integers n. By applying induction, one could prove the integrity of the given recursive formulas for creating a whole sequence from just the first term and the defined rule.

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Most popular questions from this chapter

Consider the sequence \(\left(a_{n}\right)_{n \geq 1}\) that starts \(1,3,5,7,9, \ldots\) (i.e., the odd numbers in order). (a) Give a recursive definition and closed formula for the sequence. (b) Write out the sequence \(\left(b_{n}\right)_{n \geq 2}\) of partial sums of \(\left(a_{n}\right)\). Write down the recursive definition for \(\left(b_{n}\right)\) and guess at the closed formula.

You will prove that the Fibonacci numbers satisfy the identity \(F_{n}^{2}+\) \(F_{n+1}^{2}=F_{2 n+1} .\) One way to do this is to prove the more general identity, $$ F_{m} F_{n}+F_{m+1} F_{n+1}=F_{m+n+1} $$ and realize that when \(m=n\) we get our desired result. Note that we now have two variables, so we want to prove this for all \(m \geq 0\) and all \(n \geq 0\) at the same time. For each such pair \((m, n),\) let \(P(m, n)\) be the statement \(F_{m} F_{n}+F_{m+1} F_{n+1}=F_{m+n+1}\) (a) First fix \(m=0\) and give a proof by mathematical induction that \(P(0, n)\) holds for all \(n \geq 0 .\) Note this proof will be very easy. (b) Now fix an arbitrary \(n\) and give a proof by strong mathematical induction that \(P(m, n)\) holds for all \(m \geq 0\). (c) You can now conclude that \(P(m, n)\) holds for all \(m, n \geq 0 .\) Do you believe that? Explain why this sort of induction is valid. For example, why do your proofs above guarantee that \(P(2,3)\) is true?

Make up sequences that have (a) \(3,3,3,3, \ldots\) as its second differences. (b) \(1,2,3,4,5, \ldots\) as its third differences. (c) \(1,2,4,8,16, \ldots\) as its 100 th differences.

Consider bit strings with length \(l\) and weight \(k\) (so strings of \(l 0^{\prime} \mathrm{s}\) and \(1^{\prime}\) s, including \(k 1^{\prime}\) s). We know how to count the number of these for a fixed \(l\) and \(k .\) Now, we will count the number of strings for which the sum of the length and the weight is fixed. For example, let's count all the bit strings for which \(l+k=11\). (a) Find examples of these strings of different lengths. What is the longest string possible? What is the shortest? (b) How many strings are there of each of these lengths. Use this to count the total number of strings (with sum 11 ). (c) The other approach: Let \(n=l+k\) vary. How many strings have sum \(n=1\) ? How many have sum \(n=2 ?\) And so on. Find and explain a recurrence relation for the sequence \(\left(a_{n}\right)\) which gives the number of strings with sum \(n\). (d) Describe what you have found above in terms of Pascal's Triangle. What pattern have you discovered?

Consider the sum \(4+11+18+25+\cdots+249\) (a) How many terms (summands) are in the sum? (b) Compute the sum using a technique discussed in this section.
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