Question

Give the first five terms of the given sequence. $$\left\\{a_{n}\right\\}=\left\\{\frac{4^{n}}{(n+1) !}\right\\}$$

Short answer

The first five terms are 1, 2, \(\frac{8}{3}\), \(\frac{8}{9}\), and \(\frac{16}{15}\).

Step-by-step solution

Understand the Sequence

The sequence is defined by the formula \(a_n = \frac{4^n}{(n+1)!}\). To find the terms of the sequence, you substitute different values of \(n\) starting from 0.

Calculate the First Term

Substitute \(n = 0\) into the formula: \[a_0 = \frac{4^0}{(0+1)!} = \frac{1}{1} = 1.\]So, the first term \(a_0\) is 1.

Calculate the Second Term

Substitute \(n = 1\) into the formula: \[a_1 = \frac{4^1}{(1+1)!} = \frac{4}{2!} = \frac{4}{2} = 2.\]So, the second term \(a_1\) is 2.

Calculate the Third Term

Substitute \(n = 2\) into the formula: \[a_2 = \frac{4^2}{(2+1)!} = \frac{16}{3!} = \frac{16}{6} = \frac{8}{3}.\]So, the third term \(a_2\) is \(\frac{8}{3}\).

Calculate the Fourth Term

Substitute \(n = 3\) into the formula: \[a_3 = \frac{4^3}{(3+1)!} = \frac{64}{4!} = \frac{64}{24} = \frac{8}{3}.\]So, the fourth term \(a_3\) is \(\frac{8}{9}\).

Calculate the Fifth Term

Substitute \(n = 4\) into the formula: \[a_4 = \frac{4^4}{(4+1)!} = \frac{256}{5!} = \frac{256}{120} = \frac{16}{15}.\]So, the fifth term \(a_4\) is \(\frac{16}{15}\).

Key concepts

Factorials

Factorials are an important concept in mathematics, especially useful in sequences and combinatorics. The symbol for a factorial is an exclamation mark, like this:

• The factorial of a non-negative integer, say \(n\), is denoted as \(n!\).
• It represents the product of all positive integers less than or equal to \(n\).

For instance, \(4!\) is equal to \(4 \times 3 \times 2 \times 1 = 24\). The factorial for 0, \(0!\), is defined as 1 due to the mathematical elegance it brings to various equations, including those involving permutations and series. Factorials rapidly produce large numbers, which is why sequences involving factorials tend to grow or shrink significantly as you progress through terms. In our exercise, the sequence is influenced by the denominator \((n+1)!\), demonstrating how factorials affect the size of sequence terms.

Exponential Functions

Exponential functions constitute a critical part of mathematical analysis and describe highly swift growth patterns. They take the form \(b^n\), where \(b\) is a constant base, and \(n\) is the variable exponent:

• These functions grow increasingly faster as the value of the exponent, \(n\), increases.
• This quick escalation is particularly noticeable when the base, such as \(b = 4\), is greater than 1.

In the given sequence, the term \(4^n\) demonstrates exponential growth, initially swelling the value of each term quite rapidly. However, when combined with factorials in the denominator, the sequence reflects a complex interplay. The exponential growth is countered by the factorial's growth in the denominator, influencing the nature and size of the sequence terms as \(n\) increases.

Sequence Terms

Sequence terms form the foundation of any sequence and are derived from a specific formula. Each term is often a result of substituting successive integers for a variable, like \(n\), in the formula:

• The sequence in the exercise is given by \(a_n = \frac{4^n}{(n+1)!}\), where each \(a_n\) represents a term in this sequence.
• The first five sequence terms are calculated by substituting \(n = 0, 1, 2, 3, 4\).

Calculating sequence terms involves both understanding their mathematical formulation and performing step-by-step arithmetic. For instance, starting with \(a_0\) by substituting \(n = 0\), we get 1. Each subsequent term requires recalculating based on the specified formula for changing values of \(n\), illustrating how terms evolve within the sequence's structure.

Calculus

While calculus may not seem directly connected to the specific calculation of sequence terms, understanding sequences is essential in calculus, especially in the analysis of series and limits. Here are some connections:

• Calculus helps in understanding the behavior of sequences as \(n\) approaches infinity, specifically through the concept of limits.
• Series, which are sums of sequences, are a primary focus in calculus, leading to applications in integration and differentiation.

In our exercise's sequence, understanding how the terms diminish or grow as \(n\) increases is enhanced by an appreciation of calculus principles. This enables predictions and deeper comprehension of sequence behavior beyond simple arithmetic sequences, providing insights into convergence, divergence, and other calculus-related concepts.