Problem 5 - Thomas Calculus

Question

Each of Exercises \(1-6\) gives a formula for the \(n\) th term \(a_{n}\) of a sequence \(\left\\{a_{n}\right\\} .\) Find the values of \(a_{1}, a_{2}, a_{3},\) and \(a_{4} .\) $$ a_{n}=\frac{2^{n}}{2^{n+1}} $$

Step-by-Step Solution

Answer

All terms \(a_1, a_2, a_3,\) and \(a_4\) are \(\frac{1}{2}\).

Step 1: Understand the Formula

The given formula for the sequence is \(a_{n} = \frac{2^n}{2^{n+1}}\). This formula defines how each term in the sequence is calculated based on the index \(n\).

Step 2: Simplify the Formula

Simplify the expression \(\frac{2^n}{2^{n+1}}\). By the properties of exponents, this can be written as \(\frac{2^n}{2^n \cdot 2^1} = \frac{1}{2}\). Thus, \(a_n = \frac{1}{2}\).

Step 3: Calculate \(a_1\)

Substitute \(n = 1\) into the simplified formula: \(a_1 = \frac{1}{2}\). Thus, the first term is \(\frac{1}{2}\).

Step 4: Calculate \(a_2\)

Substitute \(n = 2\) into the formula: \(a_2 = \frac{1}{2}\). The second term is also \(\frac{1}{2}\).

Step 5: Calculate \(a_3\)

Substitute \(n = 3\) into the formula: \(a_3 = \frac{1}{2}\). The third term is \(\frac{1}{2}\).

Step 6: Calculate \(a_4\)

Substitute \(n = 4\) into the formula: \(a_4 = \frac{1}{2}\). The fourth term is \(\frac{1}{2}\).

Key Concepts

nth term

The "nth term" is a fundamental idea in sequences, which helps you determine any term within the sequence. Each position in the sequence is associated with a number, starting from 1, called the "index," denoted by \( n \). The formula for the nth term, \( a_n \), in this exercise is \( \frac{2^n}{2^{n+1}} \). This formula provides you a systematic way to find any term by inserting the index number into the formula. Understanding the nth term is crucial because it eliminates the need to list out all previous terms when looking for a specific term.

• If you need \( a_1 \), you substitute \( n = 1 \) into the formula.
• For \( a_2 \), use \( n = 2 \) and so on.

This concept underlies the power of sequences in simplifying complex problems into manageable steps.

simplifying expressions

Simplifying expressions is a pivotal part of working with sequences and algebra in general. It involves reducing expressions to their simplest form to make calculations easier. In the given exercise, the sequence term \( a_n = \frac{2^n}{2^{n+1}} \) can seem complex at first. However, using properties of exponents helps simplify it.
Initially, the expression involves dividing two powers of 2. According to the rules of exponents, \( \frac{2^n}{2^{n+1}} = \frac{2^n}{2^n \cdot 2^1} \), which simplifies to \( \frac{1}{2} \).
The ability to simplify expressions is essential, as it allows you to focus on the broader problems without getting bogged down by intricate calculations. By developing skills in simplifying, you enhance your efficiency in solving mathematical problems.

exponents

Exponents are a mathematical shorthand indicating how many times a number, called the base, is multiplied by itself. In the sequence formula \( a_n = \frac{2^n}{2^{n+1}} \), you see exponents in action with the base of 2. Understanding some of the key rules will aid in simplifying such expressions:

• Product of powers: When multiplying with the same base, add the exponents, i.e., \( a^m \cdot a^n = a^{m+n} \).
• Quotient of powers: When dividing with the same base, subtract the exponents, i.e., \( \frac{a^m}{a^n} = a^{m-n} \).
• Power of a power: Multiply the exponents, i.e., \( (a^m)^n = a^{m\cdot n} \).

These facts allow us to understand why \( \frac{2^n}{2^{n+1}} \) simplifies to \( \frac{1}{2} \). Recognizing and applying these rules can dramatically reduce the complexity of problems involving powers.

calculating terms

Calculating terms in a sequence means finding the actual numbers represented by each position in the sequence. With a simplified formula for \( a_n \) from our exercise, which turns out to be \( \frac{1}{2} \) after simplification, calculating each term becomes straightforward. For any index \( n \), substituting that index into the formula \( a_n = \frac{1}{2} \) always yields \( \frac{1}{2} \).

• Calculate \( a_1 \): \( n = 1 \rightarrow a_1 = \frac{1}{2} \).
• Calculate \( a_2 \): \( n = 2 \rightarrow a_2 = \frac{1}{2} \).
• Calculate \( a_3 \): \( n = 3 \rightarrow a_3 = \frac{1}{2} \).
• Calculate \( a_4 \): \( n = 4 \rightarrow a_4 = \frac{1}{2} \).

Calculating terms this way ensures accuracy and speed, making sequences a very effective way to handle logical progressions of numbers.