Problem 76 - Precalculus with Limits

Question

In Exercises 71-76, write the first five terms of the sequence. (Assume that \( n \) begins with 0.) \( a_n = \dfrac{(-1)^{2n + 1}}{(2n + 1)!} \)

Step-by-Step Solution

Answer

The first five terms of the sequence are: -1, 1/6, -1/120, 1/5040, -1/362880.

Step 1: Calculate the first term of the sequence (n=0)

Substitute n = 0 into the sequence formula: \(a_0 = \dfrac{(-1)^{2.0 + 1}}{(2.0 + 1)!} = \dfrac{-1}{1!} = -1\)

Step 2: Calculate the second term of the sequence (n=1)

Substitute n = 1 into the sequence formula: \(a_1 = \dfrac{(-1)^{2.1 + 1}}{(2.1 + 1)!} = \dfrac{1}{3!} = \dfrac{1}{6}\)

Step 3: Calculate the third term of the sequence (n=2)

Substitute n = 2 into the sequence formula: \(a_2 = \dfrac{(-1)^{2.2 + 1}}{(2.2 + 1)!} = \dfrac{-1}{5!} = -\dfrac{1}{120}\)

Step 4: Calculate the fourth term of the sequence (n=3)

Substitute n = 3 into the sequence formula: \(a_3 = \dfrac{(-1)^{2.3 + 1}}{(2.3 + 1)!} = \dfrac{1}{7!} = \dfrac{1}{5040}\)

Step 5: Calculate the fifth term of the sequence (n=4)

Substitute n = 4 into the sequence formula: \(a_4 = \dfrac{(-1)^{2.4 + 1}}{(2.4 + 1)!} = \dfrac{-1}{9!} = -\dfrac{1}{362880}\)

Key Concepts

Factorial

Factorial is a fundamental concept in mathematics, used to represent the product of all positive integers up to a given number. It's denoted by an exclamation mark, such as \( n! \). For example:

• \( 3! = 3 \times 2 \times 1 = 6 \)
• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

The factorial grows rapidly as \( n \) increases. In the context of sequences, it often appears in denominators, as seen in our exercise, to adjust the size of each term, leading to rapid changes. Understanding factorials helps in calculating permutations and combinations where order matters. In more advanced mathematics, they play a key role in expansions like Taylor series.

Alternating Series

An alternating series is a sequence whose terms alternate in sign. Typically, it follows a pattern of positive, negative, positive, and so on. This can be represented by a factor of \((-1)^n\), which influences the sign:

• \((-1)^0 = 1\) gives a positive term.
• \((-1)^1 = -1\) gives a negative term.

In our exercise, the sequence is \( a_n = \frac{(-1)^{2n + 1}}{(2n + 1)!} \). The exponent \(2n + 1\) ensures that each term alternates. Alternating series are significant in mathematical analysis and can help in estimating functions and ensuring convergence. They often appear in the study of trigonometric and logarithmic functions.

Calculating Terms

Calculating terms in a sequence involves substituting values into a given formula. Starting with \( n = 0 \), you substitute and simplify:

• For \( n = 0 \): \( a_0 = \frac{(-1)^{2 \times 0 + 1}}{(2 \times 0 + 1)!} = -1 \)
• For \( n = 1 \): \( a_1 = \frac{(-1)^{2 \times 1 + 1}}{(2 \times 1 + 1)!} = \frac{1}{6} \)
• Continue with \( n = 2, 3, 4 \) to find \( a_2, a_3, a_4 \).

Breaking down each step clarifies how changes in \( n \) affect the term's value. This specific sequence formula involves both an alternating component \((-1)\) and a factorial in the denominator, which influences the magnitude. Understanding these elements allows for accurate computation of terms, essential for analyzing the sequence's behavior.