Problem 29 - Algebra and Trigonometry

Question

The given pattern continues. Write down the nth term of a sequence \(\left\\{a_{n}\right\\}\) suggested by the pattern. \(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots\)

Step-by-Step Solution

Answer

\(a_n = 2^{-(n-1)}\)

Step 1 - Identify the pattern

Observe the given sequence: 1, \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), \ldots. Notice that each term seems to be half of the previous term.

Step 2 - Express each term

Write down the first few terms with exponents: \(1 = 2^0\), \(\frac{1}{2} = 2^{-1}\), \(\frac{1}{4} = 2^{-2}\), \(\frac{1}{8} = 2^{-3}\). This shows that the exponents follow the pattern of increasing negative integers.

Step 3 - Generalize the nth term

Recognize that the exponent in each term corresponds to \(-(n-1)\). Therefore, the nth term can be written as \(a_n = 2^{-(n-1)}\).

Key Concepts

geometric sequence

A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In this exercise, the given sequence is: 1, \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{8}\), ... Notice that each term is half of the term before it. This means our common ratio is \(\frac{1}{2}\). Understanding geometric sequences is key to solving problems where patterns involve multiplication or division by a constant factor.
Identify the common ratio:

• First term (\(a\)): 1
• Common ratio (\(r\)): \(\frac{1}{2}\)

Here, each term can be represented as \(a_n = a \times r^{(n-1)}\). Since our first term (\(a\)) is 1 and our common ratio (\(r\)) is \(\frac{1}{2}\), this formula will help us find any term in our sequence.

exponents

Exponents indicate how many times a number, called the base, is multiplied by itself. For instance, in the term \(2^3\), 2 is the base and 3 is the exponent, indicating 2 is multiplied by itself 3 times, resulting in 8. In this exercise, the sequence involves negative exponents. Let’s break down the terms provided:

• \(1 = 2^0\)
• \(\frac{1}{2} = 2^{-1}\)
• \(\frac{1}{4} = 2^{-2}\)
• \(\frac{1}{8} = 2^{-3}\)

Each subsequent term has the exponent decreasing by 1. Negative exponents indicate reciprocals: \(2^{-n} = \frac{1}{2^n}\). This is essential for understanding terms in sequences like the geometric sequence in this exercise.

nth term formula

The nth term formula is a way to describe the term at any position in a sequence using a mathematical expression. It allows us to find any term without listing all prior terms. For the geometric sequence in this exercise, we have determined that the common ratio \(r\) is \(\frac{1}{2}\) and the first term \(a\) is 1.
We saw that each term is expressed using negative exponents:

• \(1 = 2^0\)
• \(\frac{1}{2} = 2^{-1}\)
• \(\frac{1}{4} = 2^{-2}\)
• \(\frac{1}{8} = 2^{-3}\)

This gives us the nth term formula: \(a_n = 2^{-(n-1)}\). Using this formula, we can find any term in the sequence without listing all previous terms. For example, the 5th term is: \[a_5 = 2^{-(5-1)} = 2^{-4} = \frac{1}{16}\]