Question

Find an explicit formula for the \(n\) th term of the sequence satisfying \(a_{1}=0\) and \(a_{n}=2 a_{n-1}+1\) for \(n \geq 2\).

Short answer

The explicit formula is \(a_n = 2^{n-1} - 1\).

Step-by-step solution

Understanding the Recurrence Relation

We start by looking at the given recurrence relation. We know that the sequence is defined recursively as \(a_{n} = 2a_{n-1} + 1\), with an initial condition of \(a_1 = 0\). This sequence builds on its previous terms.

Writing Down Initial Terms

To make sense of the sequence, we calculate the first few terms:- For \(n = 1\), \(a_1 = 0\).- For \(n = 2\), \(a_2 = 2 \cdot a_1 + 1 = 1\).- For \(n = 3\), \(a_3 = 2 \cdot a_2 + 1 = 3\).- For \(n = 4\), \(a_4 = 2 \cdot a_3 + 1 = 7\).The sequence appears to be growing rapidly.

Identifying a Pattern

Let's observe the pattern in the sequence by comparing the terms:- \(a_1 = 0\)- \(a_2 = 1 = 2^1 - 1\)- \(a_3 = 3 = 2^2 - 1\)- \(a_4 = 7 = 2^3 - 1\)We notice that each term seems to follow the pattern \(a_n = 2^{n-1} - 1\).

Verifying the Pattern

We need to verify that the pattern \(a_n = 2^{n-1} - 1\) satisfies the given recurrence relation.1. Base case: - For \(n = 1\), \(a_1 = 2^{1-1} - 1 = 0\). This matches the initial condition.2. Inductive step: - Assume \(a_k = 2^{k-1} - 1\) is true. - Show \(a_{k+1} = 2^k - 1\):Replace \(a_k\) in the recurrence: - \(a_{k+1} = 2a_k + 1\) - \(a_{k+1} = 2(2^{k-1} - 1) + 1 = 2^k - 2 + 1 = 2^k - 1\).This confirms the pattern holds for all \(n\).

Concluding the Explicit Formula

Thus, the explicit formula for the \(n\)th term of the sequence is: \[ a_n = 2^{n-1} - 1 \]

Key concepts

explicit formula

In mathematics, an explicit formula provides a direct way to calculate the nth term of a sequence without needing to refer to previous terms. For our sequence defined by the recurrence relation \( a_{n} = 2a_{n-1} + 1 \) with the initial term \( a_1 = 0 \), we can simplify our work by forming such a formula.

Once we identify the pattern, we find an explicit formula which expresses \( a_n \) as a function of \( n \). This means, instead of calculating every term based on the previous one, we can plug \( n \) into an easy-to-manage formula. Our step-by-step work reveals that the sequence pattern adheres to \( a_n = 2^{n-1} - 1 \).

This formula takes advantage of properties of powers of 2. Specifically, each new term in the sequence is derived as 2 raised to the power of one less than the term’s position, then subtracting one. An explicit formula such as this is very useful because it saves us time and computation effort for large \( n \), providing instant access to any term in the sequence.

inductive reasoning

Inductive reasoning is a process of reasoning in which generalizations are drawn from specific examples. It starts with specific observations, and from these, general conclusions are made. In the context of sequences and recurrence relations, inductive reasoning helps verify that a proposed formula holds true for all terms.

To confirm our explicit formula \( a_n = 2^{n-1} - 1 \), we use the mathematical principle of induction. First, we check that the base case satisfies the formula. For \( n = 1 \), \( a_1 = 2^{0} - 1 = 0 \), which matches our initial condition.
Then, we assume that the formula works for a term \( a_k \), so \( a_k = 2^{k-1} - 1 \). The next step, known as the inductive step, is to demonstrate it connects to \( a_{k+1} \). That means showing \( a_{k+1} = 2^k - 1 \) using the formula \( a_{k+1} = 2a_k + 1 \).
By replacing \( a_k \) with \( 2^{k-1} - 1 \) and performing the calculation, we see that both sides match. This confirms our pattern holds for all consecutive terms and validates the general formula.

sequence pattern

Understanding sequence patterns is key to analyzing sequences defined by recurrence relations. Patterns reveal important characteristics that guide us to finding an explicit formula.

For the sequence given by \( a_{n} = 2a_{n-1} + 1 \), we began by calculating initial values to spot a pattern. The pattern emerges as:

• \( a_1 = 0 \)
• \( a_2 = 1 = 2^1 - 1 \)
• \( a_3 = 3 = 2^2 - 1 \)
• \( a_4 = 7 = 2^3 - 1 \)

From these terms, the pattern \( a_n = 2^{n-1} - 1 \) is recognized. Standard patterns in sequences often involve powers, factors, or recognizable numbers such as squares or cubes.
Detecting these patterns allows us to leverage these observations to derive explicit formulas and better understand the sequences. Whether we are creating or analyzing sequences, noticing and confirming these patterns is crucial for determining the sequence's long-term behavior and for simplifying calculations.