Problem 99 - Thomas Calculus

Question

The first term of a sequence is \(x_{1}=1 .\) Each succeeding term is the sum of all those that come before it: $$x_{n+1}=x_{1}+x_{2}+\dots+x_{n}$$ Write out enough early terms of the sequence to deduce a general formula for \(x_{n}\) that holds for \(n \geq 2\) .

Step-by-Step Solution

Answer

The general formula for the sequence is \(x_n = 2^{n-2}\) for \(n \geq 2\).

Step 1: Write the Current Terms

Start by writing down the first term of the sequence. We have \(x_1 = 1\).

Step 2: Calculate the Second Term

The second term is the sum of all terms before it, which is only \(x_1\) since \(x_2 = x_1 = 1\). Thus, \(x_2 = 1\).

Step 3: Calculate the Third Term

The third term is the sum of the first and second terms, so \(x_3 = x_1 + x_2 = 1 + 1 = 2\).

Step 4: Calculate the Fourth Term

Continuing the pattern, the fourth term is the sum of the first, second, and third terms. Thus, \(x_4 = x_1 + x_2 + x_3 = 1 + 1 + 2 = 4\).

Step 5: Calculate the Fifth Term

The fifth term is the sum of all terms before it: \(x_5 = x_1 + x_2 + x_3 + x_4 = 1 + 1 + 2 + 4 = 8\).

Step 6: Recognize the Pattern

Notice that each term, starting from \(x_2\), can be expressed as powers of 2: \(x_2 = 1\), \(x_3 = 2\), \(x_4 = 4 = 2^2\), \(x_5 = 8 = 2^3\). This suggests that \(x_n = 2^{n-2}\) for \(n \geq 2\).

Step 7: General Formula Confirmation

Based on the pattern, the formula for the \(n^{th}\) term is \(x_n = 2^{n-2}\) for \(n \geq 2\). Verify this with calculations found in previous steps for confirmation.

Key Concepts

Recursive Sequences

Recursive sequences are a fascinating area of mathematics where each term depends on the previous ones in some way. In our exercise, the sequence definition is recursive because each term is the sum of all preceding terms. The process begins with a known starting term, the base case, from which subsequent terms are calculated. In this case, the first term is given as \(x_1 = 1\). In general, a recursive sequence can be defined by a formula where a term \(x_{n+1}\) is expressed in terms of previous terms. For example, a recursive definition may look like \(x_{n+1} = f(x_n)\), where \(f\) is some function. In our case, \(f(x)\) is the sum function, accumulating all terms up to \(x_n\). Recursive sequences can illustrate complex patterns and behaviors, making them an intriguing subject for mathematical exploration.

Series and Patterns

Series and patterns are at the heart of identifying rules governing sequences. In our sequence, we noticed that each term doubles the previous term once we move past the initial terms. This doubling pattern is key to recognizing that each term can be expressed as powers of two.To illustrate:

• \(x_2 = 1 = 2^0\)
• \(x_3 = 2 = 2^1\)
• \(x_4 = 4 = 2^2\)
• \(x_5 = 8 = 2^3\)

Identifying such series and patterns enables us to establish a relationship or formula for predicting any term in the sequence. It's fascinating how simple observations evolve into structured and predictable mathematical expressions, providing insights into how sequences grow.

Mathematical Induction

Mathematical induction is a powerful technique used to prove the validity of statements or formulas related to sequences, often applied when patterns are established. In our exercise, mathematical induction helps ensure that the formula \(x_n = 2^{n-2}\) is valid for all \(n \geq 2\).The method involves two main steps:- **Base Case:** Verify the statement for the initial value, \(n = 2\), where \(x_2 = 2^{2-2} = 1\), which matches the sequence.- **Inductive Step:** Assume the formula holds for \(n = k\), and then prove it true for \(n = k+1\). This means if the assertion holds for \(n = k\), then \(x_{k+1} = 2^{(k+1)-2} = 2^{k-1}\) must also be true.If both steps are successful, the formula is confirmed true for all relevant \(n\), offering a solid proof method. Mathematical induction can handle complex sequence problems efficiently by proving propositions stage-by-stage in a straightforward manner.