Question

Find a function \(f(n)\) that identifies the \(n\) th term \(a_{n}\) of the following recursively defined sequences, as \(a_{n}=f(n)\). $$ a_{1}=1 \text { and } a_{n+1}=(n+1) a_{n} \text { for } n \geq 1 $$

Short answer

The function is \(f(n) = n!\).

Step-by-step solution

Understanding the Problem

We have a sequence defined recursively: the first term is \(a_1 = 1\) and each subsequent term is defined as \(a_{n+1} = (n+1) \, a_n\). We need to express the \(n\)-th term \(a_n\) as a function \(f(n)\).

Recognizing the Pattern

From the recursive definition, we can deduce a pattern. The sequence is generated by multiplying the previous term by \(n+1\). This is characteristic of a factorial, where \(a_n = n!\).

Deriving the Function

To find a general formula, start with initial terms: \(\begin{align*} a_1 &= 1, \ a_2 &= 2 \times a_1 = 2 \ a_3 &= 3 \times a_2 = 6 \ a_4 &= 4 \times a_3 = 24. \end{align*}\) The pattern suggests that \(a_n = n!\).

Verifying the Function

We verify by induction: Assume \(a_n = n!\) is true, then for \(a_{n+1} = (n+1)\cdot a_n = (n+1)\cdot n! = (n+1)!\). This confirms our function is \(f(n) = n!\).

Writing the Function

The function that gives the \(n\)-th term of the sequence is \(f(n) = n!\). This compactly represents the behavior and definition of the sequence.

Key concepts

Factorial Sequence

The factorial sequence is a sequence of numbers in which each number is the product of all positive integers less than or equal to that number. For example, the factorial of a number denoted as \( n! \) is calculated as \( n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1 \).
This concept is central to many areas of mathematics, particularly in combinatorics, as it describes the total arrangements of a set of items.
In our problem, we are asked to find each term of a recursively defined sequence, and we discovered that each term is indeed a factorial. Here, the first term is \( a_1 = 1 \). Each subsequent term is formed by multiplying the current index with the previous term, which matches the pattern of calculating a factorial.

• Starting with \( a_1 = 1 \)
• Continuing with \( a_2 = 2 \times 1 = 2 \)
• Then \( a_3 = 3 \times 2 = 6 \), and so on

This led us to conclude that \( a_n = n! \) for each term.

Recursive Sequence

A recursive sequence is a sequence in which terms are defined in terms of previous terms. This means each subsequent term follows the same pattern or rule.
The sequence is initialized with one or more starting values, and a recursive formula defines each term as a function of the preceding term(s). In our exercise, the sequence is given by the recursive formula \( a_{n+1} = (n+1) \cdot a_n \), which means each term is computed by multiplying the previous term by one more than the term number.

• Recursion allows sequences to be defined succinctly.
• Each term builds upon the previous, creating a chain of dependencies.
• Understanding the initial condition is crucial, with \( a_1 = 1 \) here.

This recursive nature can simplify defining complex sequences elegantly instead of listing or calculating each term independently.

Mathematical Induction

Mathematical induction is a powerful proof technique used to demonstrate that a statement or formula is true for all natural numbers.
It involves two essential steps: the base case and the inductive step. In our problem, induction is used to prove that the function \( f(n) = n! \) correctly finds each term of the sequence.
First, we verify the base case: \( a_1 = 1! \), which is true. Then for our inductive step, assume \( a_n = n! \) holds true for some arbitrary \( n \).

• Show that the formula is true for \( n = 1 \).
• Assume \( a_n = n! \) is true (inductive hypothesis).
• Prove \( a_{n+1} = (n+1)! \).

Using this method, we show that our formula holds for all positive integers \( n \), proving the entire sequence formula is valid and exact.