Problem 78 - Precalculus: Mathematical for Calculus

Question

Different Sequences That Start the Same (a) Show that the first four terms of the sequence \(a_{n}=n^{2}\) are $$ 1,4,9,16, \dots $$ (b) Show that the first four terms of the sequence $$ \begin{array}{c} a_{n}=n^{2}+(n-1)(n-2)(n-3)(n-4) \text { are also } \\ 1,4,9,16, \ldots \end{array} $$ (c) Find a sequence whose first six terms are the same as those of \(a_{n}=n^{2}\) but whose succeeding terms differ from this sequence. (d) Find two different sequences that begin $$ 2,4,8,16, \dots $$

Step-by-Step Solution

Answer

Sequences a: 1,4,9,16; b: same. c) New: add (n-1)(n-2)(n-3)(n-4)(n-5). d) Examples: 2^n, recursive doubling.

Step 1: First Terms of Sequence (a)

Given the sequence \( a_n = n^2 \), for \( n = 1, 2, 3, 4 \), calculate the terms:- For \( n = 1 \), \( a_1 = 1^2 = 1 \)- For \( n = 2 \), \( a_2 = 2^2 = 4 \)- For \( n = 3 \), \( a_3 = 3^2 = 9 \)- For \( n = 4 \), \( a_4 = 4^2 = 16 \)Thus, the first four terms are 1, 4, 9, 16.

Step 2: First Terms of Sequence (b)

Given the sequence \( a_n = n^2 + (n-1)(n-2)(n-3)(n-4) \), calculate the first four terms:- For \( n=1 \) to \( n=4 \), the term \( (n-1)(n-2)(n-3)(n-4) = 0 \) (since one of the factors is zero)Hence,- \( a_1 = 1^2 + 0 = 1 \)- \( a_2 = 2^2 + 0 = 4 \)- \( a_3 = 3^2 + 0 = 9 \)- \( a_4 = 4^2 + 0 = 16 \)The first four terms are the same: 1, 4, 9, 16.

Step 3: Finding a New Sequence for Part (c)

We need a sequence matching \( n^2 \) for the first 6 terms but differing thereafter. Consider \[ a_n = n^2 + (n-1)(n-2)(n-3)(n-4)(n-5) \]For \( n=1 \) to \( n=6 \), the additional term is zero, matching \( n^2 \):- \( a_1 = 1, a_2 = 4, a_3 = 9, a_4 = 16, a_5 = 25, a_6 = 36 \)For \( n = 7 \), it becomes \( a_7 = 49 + 720 = 769 \), differing from 49.

Step 4: Sequence Starting at 2, 4, 8, 16 for Part (d)

To find two sequences starting 2, 4, 8, 16:1. Consider a geometric sequence: \( a_n = 2^n \). - For \( n = 1 \), \( a_1 = 2^1 = 2 \) - For \( n = 2 \), \( a_2 = 2^2 = 4 \) - For \( n = 3 \), \( a_3 = 2^3 = 8 \) - For \( n = 4 \), \( a_4 = 2^4 = 16 \)2. Consider the recursive sequence: start with 2, then \( a_{n} = 2a_{n-1} \). - \( a_1 = 2 \), \( a_2 = 4 \), \( a_3 = 8 \), \( a_4 = 16 \)

Key Concepts

Geometric Sequence

A geometric sequence is a type of sequence where each term is found by multiplying the previous term by a constant, called the ratio. This ratio remains the same throughout the sequence.
In the given exercise, one example of this is the sequence starting with 2, 4, 8, 16. To understand this more simply, let's break it down:

• The first term, often called the initial term, is 2.
• To find the next term, you multiply the current term by the common ratio, which in this sequence is 2.
• So, starting at 2, the sequence develops as follows: multiply 2 by 2 to get 4, then multiply 4 by 2 to get 8, and so on.

Notice how every time you find the next number, you're just doubling the one before it. It's like a repeating pattern that grows exponentially. By understanding this method, you can quickly identify or construct any geometric sequence, making it an essential concept for recognizing patterns in mathematics.

Recursive Sequence

A recursive sequence is a sequence where each term is defined in relation to the term(s) before it. This is like using the outcome of one step as the starting point for the next.
In the step-by-step solution, a recursive approach is shown for the sequence starting 2, 4, 8, 16 by using a simple rule: each term is double the previous term. Let's explore this:

• You start with the initial term, which is known beforehand — here it is 2.
• Each subsequent term is defined based on the prior term, often via a defined equation.

For example, if you define a recursive sequence as \( a_n = 2a_{n-1} \), and you know \( a_1 = 2 \), you can find \( a_2 \) by calculating \( 2 \times a_1 = 2 \times 2 = 4 \), and then \( a_3 = 2 \times a_2 = 8 \), and so forth.
This structure of defining terms based on others is what makes it recursive, allowing problems to be solved incrementally, term by term.

Quadratic Sequence

Quadratic sequences are sequences where the differences between terms move quadratically. Essentially, each number is related to the square of the term's position in the sequence.
The simplest example is the sequence \( a_n = n^2 \) that appears in the exercise and solutions:

• The first term \( n = 1 \) is squared to get \( 1^2 = 1 \).
• The second term \( n = 2 \) becomes \( 2^2 = 4 \).
• Following this pattern, \( n = 3 \) becomes \( 9 \), and \( n = 4 \) becomes \( 16 \).

Each term is simply its position squared, creating a simple yet powerful quadratic progression. Understanding quadratic sequences helps in recognizing how sequences grow and change, capturing the essence of squared growth, a concept very fundamental in different areas of math, such as algebra and calculus.