Problem 6 - 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 \(a_{4} .\) $$ a_{n}=\frac{2^{n}-1}{2^{n}} $$

Step-by-Step Solution

Answer

\( a_1 = \frac{1}{2}, a_2 = \frac{3}{4}, a_3 = \frac{7}{8}, a_4 = \frac{15}{16} \)

Step 1: Calculate a_1

To find the first term of the sequence, substitute \( n = 1 \) into the formula \( a_n = \frac{2^n - 1}{2^n} \). This gives: \[ a_1 = \frac{2^1 - 1}{2^1} = \frac{2 - 1}{2} = \frac{1}{2} \] So, the value of \( a_1 \) is \( \frac{1}{2} \).

Step 2: Calculate a_2

Substitute \( n = 2 \) into the formula \( a_n = \frac{2^n - 1}{2^n} \) to find the second term: \[ a_2 = \frac{2^2 - 1}{2^2} = \frac{4 - 1}{4} = \frac{3}{4} \] Thus, \( a_2 = \frac{3}{4} \).

Step 3: Calculate a_3

Substitute \( n = 3 \) into the formula \( a_n = \frac{2^n - 1}{2^n} \) to find the third term: \[ a_3 = \frac{2^3 - 1}{2^3} = \frac{8 - 1}{8} = \frac{7}{8} \] Thus, \( a_3 = \frac{7}{8} \).

Step 4: Calculate a_4

Substitute \( n = 4 \) into the formula \( a_n = \frac{2^n - 1}{2^n} \) to find the fourth term: \[ a_4 = \frac{2^4 - 1}{2^4} = \frac{16 - 1}{16} = \frac{15}{16} \] Thus, \( a_4 = \frac{15}{16} \).

Key Concepts

Arithmetic Sequences

An arithmetic sequence is a list of numbers with a definite pattern. Each number is called a term. In an arithmetic sequence, the difference between consecutive terms is always the same, known as the "common difference." This is one of the simplest sequences you can see in mathematics.

Here’s how it works:

• Start with the first term, usually denoted as \( a_1 \).
• Add the common difference \( d \) to get the next term.
• Repeat this addition to generate more terms.

For example, consider the sequence \( 2, 5, 8, 11, \ldots \). Here, the common difference \( d \) is 3 because you add 3 to get from one term to the next.

The general formula for the nth term of an arithmetic sequence is:
\[ a_n = a_1 + (n-1) \cdot d \] Using this formula, you can find any term in the sequence by knowing the first term \( a_1 \), the common difference \( d \), and which term number \( n \) you want to find.

Geometric Sequences

A geometric sequence is another sequence pattern where each term is obtained by multiplying the previous term by a fixed, non-zero number called the "common ratio." This type of sequence might not seem as straightforward as arithmetic sequences, but they play a crucial role in various applications like computing interest or analyzing population growth.

Let's break it down:

• Start with the first term, \( a_1 \).
• Multiply by the common ratio \( r \) to get from one term to the next.
• Continue multiplying by \( r \) for subsequent terms.

The pattern is evident in sequences like \( 3, 6, 12, 24, \ldots \), where each term is twice the previous one, which means the common ratio \( r \) is 2.

The formula to find the nth term of a geometric sequence is:
\[ a_n = a_1 \cdot r^{n-1} \] With this formula, having the first term \( a_1 \) and the common ratio \( r \), you can easily calculate any term in the sequence, even those far out in the sequence.

Formula for nth Term

The formula for the nth term of a sequence allows you to determine any term in a sequence without listing all the previous terms. This can save a lot of time, especially if you want to find terms located far into the sequence.

There are different formulas for different kinds of sequences. Let's explore some key formulas:

• For arithmetic sequences, use \( a_n = a_1 + (n-1) \cdot d \), where \( d \) is the common difference.
• For geometric sequences, use \( a_n = a_1 \cdot r^{n-1} \), where \( r \) is the common ratio.
• If you're given a specific sequence formula, like \( a_n = \frac{2^n - 1}{2^n} \), plug the desired term number \( n \) into the equation to find that term.

Understanding these formulas allows you to work with sequences efficiently, revealing terms quickly without needing manual calculations each time. Whether you're dealing with arithmetic or geometric sequences, or even unique formulas, applying the correct nth term formula helps simplify the problem.