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Chapter 11: Problem 17

In Exercises \(13-22,\) find a formula for the \(n\) th term of the sequence. The sequence \(0,3,8,15,24, \dots\)

Short Answer

Expert verified
The formula for the nth term of the sequence is \( a_n = n^2 - 1 \).

Step by step solution

01

Analyze the Sequence

The sequence given is: 0, 3, 8, 15, 24, ... Let's look at how the sequence is formed. We will calculate the differences between consecutive terms to identify a pattern.
02

Find First Differences

Calculate the difference between consecutive terms: 3 - 0 = 3 8 - 3 = 5 15 - 8 = 7 24 - 15 = 9 The first differences are: 3, 5, 7, 9,... This is an arithmetic sequence with a common difference of 2.
03

Identify Type of Sequence

Since the first differences form an arithmetic sequence, the original sequence is a quadratic sequence. This implies the general form of the nth term should be quadratic, \[ a_n = an^2 + bn + c \] .
04

Solve for Coefficients of Quadratic

We'll use the first three terms to find the coefficients of the quadratic:- For n = 1: \[ a(1)^2 + b(1) + c = 0 \]- For n = 2: \[ a(2)^2 + b(2) + c = 3 \]- For n = 3: \[ a(3)^2 + b(3) + c = 8 \]This gives us the system of equations:a + b + c = 0,4a + 2b + c = 3,9a + 3b + c = 8.
05

Calculate Values for a, b, and c

Subtract the first equation from the second to eliminate c:\( (4a + 2b + c) - (a + b + c) = 3 \)\[ 3a + b = 3 \] (Equation 4)Subtract the second from the third to eliminate c:\( (9a + 3b + c) - (4a + 2b + c) = 5 \)\[ 5a + b = 5 \] (Equation 5)Subtract Equation 4 from Equation 5:\( (5a + b) - (3a + b) = 5 - 3 \)\[ 2a = 2 \] Thus, \( a = 1 \) .
06

Determine Remaining Coefficients

From \( 3a + b = 3 \):Substitute \( a = 1 \) into 3a + b = 3:\( 3(1) + b = 3 \)\( b = 0 \) .Substitute \( a = 1 \) and \( b = 0 \) into the first equation:\( 1 + 0 + c = 0 \)\( c = -1 \) .
07

Write the Final Formula

Substituting the values of a, b, and c into the quadratic form, \[ a_n = n^2 - 1 \] . This is the formula for the nth term of the sequence.

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Key Concepts

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

nth Term Formula
A quadratic sequence is a sequence in which the sequence of the first differences forms an arithmetic sequence. To find a formula for the nth term of a quadratic sequence, we typically use the formula: \[ a_n = an^2 + bn + c \]. Here, \(a\), \(b\), and \(c\) are constants that we need to determine.
To derive these constants, select values from the sequence and create equations to solve for \(a\), \(b\), and \(c\). In our specific sequence: 0, 3, 8, 15, 24,..., the relationship was identified as \(a_n = n^2 - 1\).
The nth term formula gives a direct way to calculate any term in the sequence without listing all previous terms. This formula is powerful because it encapsulates the entire pattern of the sequence in one expression.
Arithmetic Sequence
While exploring sequences, an arithmetic sequence catches our attention due to its uniform pattern. It is a sequence of numbers where each term after the first is obtained by adding a constant called the 'common difference' to the preceding term.
In the context of our quadratic sequence, the differences between each consecutive term: 3, 5, 7, and 9, reveal an arithmetic sequence. Here, the common difference is 2.
  • First term: 3
  • Difference between terms: 2
This revelation helped us identify the overall sequence as quadratic since the first differences form an arithmetic sequence.
Understanding this part is crucial because it helps establish whether a sequence might be quadratic. This involves noticing that when you take the first differences of a quadratic sequence, those differences will form an arithmetic sequence.
Difference Calculations
Difference Calculations serve as a fundamental step in identifying the type of sequence and how it evolves. By calculating the differences between consecutive terms, you gain insights into the sequence's nature. In a quadratic sequence, the first differences themselves form an arithmetic sequence.
For example, in the sequence: 0, 3, 8, 15, 24,..., calculating differences between each term gives: 3, 5, 7, 9,... Hey! That's an arithmetic progression! If we calculate the differences of these numbers, we notice they are constant:
  • 3 - 0 = 3
  • 5 - 3 = 2
  • 7 - 5 = 2
  • 9 - 7 = 2
This consistency in the second differences (they are all 2) confirms the sequence is quadratic. Recognizing this saves time and mental energy, making solving problems much smoother.
Quadratic Formula
The quadratic formula is a powerful tool used to find the nth term of many sequences that follow a specific pattern. For a quadratic sequence, like the one we have, the nth term is typically represented in this format: \[ a_n = an^2 + bn + c \].
To find the relationship, we need to compute the coefficients \(a\), \(b\), and \(c\) by substituting sequential values into the general formula. Systematically setting up equations through particular terms, and solving step-by-step enables us to obtain these coefficients accurately.
In our example, we established: \[ a_n = n^2 - 1 \].
This quadratic formula captures the nature of the sequence by involving the square of the term position \(n\). Importantly, the quadratic nature indicates the growth pattern involves square power, crucial in many real-world applications such as physics and engineering.

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Most popular questions from this chapter

According to a front-page article in the December \(15,1992,\) issue of the Wall Street Journal, Ford Motor Company used about 7\(\frac{1}{4}\) hours of labor to produce stampings for the average vehicle, down from an estimated 15 hours in \(1980 .\) The Japanese needed only about 3\(\frac{1}{2}\) hours. Ford's improvement since 1980 represents an average decrease of 6\(\%\) per year. If that rate continues, then \(n\) years from 1992 Ford will use about $$ S_{n}=7.25(0.94)^{n} $$ hours of labor to produce stampings for the average vehicle. Assuming that the Japanese continue to spend 3\(\frac{1}{2}\) hours per vehicle, how many more years will it take Ford to catch up? Find out two ways: a. Find the first term of the sequence \(\left\\{S_{n}\right\\}\) that is less than or equal to \(3.5 .\) b. Graph \(f(x)=7.25(0.94)^{x}\) and use Trace to find where the graph crosses the line \(y=3.5 .\)

Show that if \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) both converge absolutely, then so does $$ \begin{array}{ll}{\text { a. } \sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right)} & {\text { b. } \sum_{n=1}^{\infty}\left(a_{n}-b_{n}\right)} \\ {\text { c. } \sum_{n=1}^{\infty} k a_{n} \quad(k \text { any number })}\end{array} $$

a. Suppose that \(f(x)\) is differentiable for all \(x\) in \([0,1]\) and that \(f(0)=0 .\) Define the sequence \(\left\\{a_{n}\right\\}\) by the rule \(a_{n}=\) \(n f(1 / n) .\) Show that \(\lim _{n \rightarrow \infty} a_{n}=f^{\prime}(0)\) Use the result in part (a) to find the limits of the following sequences \(\left\\{a_{n}\right\\}\). \(\begin{array}{ll}{\text { b. } a_{n}=n \tan ^{-1} \frac{1}{n}} & {\text { c. } a_{n}=n\left(e^{1 / n}-1\right)} \\ {\text { d. } a_{n}=n \ln \left(1+\frac{2}{n}\right)}\end{array}\)

Use series to evaluate the limits in Exercises \(47-56\) $$ \lim _{x \rightarrow \infty} x^{2}\left(e^{-1 / x^{2}}-1\right) $$

Estimate the error if \(\cos \sqrt{t}\) is approximated by \(1-\frac{t}{2}+\frac{t^{2}}{4 !}-\frac{t^{3}}{6 !}\) in the integral \(\int_{0}^{1} \cos \sqrt{t} d t\)
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