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Chapter 11: Problem 489

Write the first four terms of the sequence defined by the explicit formula \(a_{n}=\frac{n^{2}-n-1}{n !}\)

Short Answer

Expert verified
The first four terms are -1, \( \frac{1}{2} \), \( \frac{5}{6} \), and \( \frac{11}{24} \).

Step by step solution

01

Understand the Explicit Formula

The sequence is defined by the explicit formula \(a_{n} = \frac{n^{2}-n-1}{n!}\), where \(n!\) denotes the factorial of \(n\). This means for each term \(a_n\), we need to substitute the value of \(n\) and calculate the expression.
02

Compute the First Term

For the first term, substitute \(n = 1\) into the formula: \[ a_1 = \frac{1^{2} - 1 - 1}{1!} = \frac{1 - 1 - 1}{1} = \frac{-1}{1} = -1 \] Therefore, the first term \(a_1 = -1\).
03

Compute the Second Term

For the second term, substitute \(n = 2\) into the formula: \[ a_2 = \frac{2^{2} - 2 - 1}{2!} = \frac{4 - 2 - 1}{2} = \frac{1}{2} \] Therefore, the second term \(a_2 = \frac{1}{2}\).
04

Compute the Third Term

For the third term, substitute \(n = 3\) into the formula: \[ a_3 = \frac{3^{2} - 3 - 1}{3!} = \frac{9 - 3 - 1}{6} = \frac{5}{6} \] Therefore, the third term \(a_3 = \frac{5}{6}\).
05

Compute the Fourth Term

For the fourth term, substitute \(n = 4\) into the formula: \[ a_4 = \frac{4^{2} - 4 - 1}{4!} = \frac{16 - 4 - 1}{24} = \frac{11}{24} \] Therefore, the fourth term \(a_4 = \frac{11}{24}\).

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Key Concepts

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

Explicit Formula
An explicit formula is a mathematical expression that allows you to calculate any term in a sequence directly, without needing to find previous terms. In the context of sequences, it provides a direct relationship between the term number, often denoted as \( n \), and the term's value, \( a_n \). This is incredibly useful because you can determine any term without calculating all prior terms, saving time and effort.
  • Often appears in the form \( a_n = f(n) \)
  • Allows you to calculate terms quickly
  • Eliminates the need for recursive calculations
In the exercise, the explicit formula \( a_n = \frac{n^2-n-1}{n!} \) is used to find specific terms of a sequence.
Factorial
The factorial of a number, denoted \( n! \), is the product of all positive integers less than or equal to \( n \). It plays a crucial role in permutations, combinations, and many areas of mathematics, including sequences like the one in the exercise.
  • Defined as \( n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \)
  • Factorials grow rapidly, making calculations larger
  • Used in the denominator of the explicit formula \( a_n = \frac{n^2-n-1}{n!} \)
For example, when \( n = 4 \), the factorial \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). The factorial part in the denominator simplifies calculations and is a critical component in determining the values of sequence terms.
Sequence Terms
In any sequence, each number is called a term, and each term can be calculated or represented using a specific formula or rule. The sequence in question begins from \( n = 1 \), and we compute the terms using their explicit formula.
  • The first term, \( a_1 \), is \(-1\)
  • The second term, \( a_2 \), is \(\frac{1}{2}\)
  • The third term, \( a_3 \), is \(\frac{5}{6}\)
  • The fourth term, \( a_4 \), is \(\frac{11}{24}\)
Each term is derived by plugging the sequence number \( n \) into the explicit formula. This shows how the terms evolve as \( n \) increases, providing insight into the behavior of the sequence.
Mathematical Expressions
Mathematical expressions are a combination of numbers, variables, and symbols to represent quantities and operations. They are the core building blocks in mathematics, allowing us to articulate and solve problems effectively.
  • Include operations like addition, subtraction, multiplication, and division
  • Used in crafting formulas such as the explicit formula
  • Enable calculation of complex problems in a structured way
The expression \( \frac{n^2-n-1}{n!} \) in this sequence involves squaring \( n \), subtracting, and then dividing by the factorial of \( n \). This complex interplay of operations showcases the power and flexibility of mathematical expressions in solving problems and describing sequences.

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Most popular questions from this chapter

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