Chapter 11: Problem 66
Find the first 4 terms of the recursively defined sequence. $$a_{1}=256, a_{n+1}=\sqrt{a_{n}}$$
Short Answer
Expert verified
256, 16, 4, 2
Step by step solution
01
Understand the problem
Identify that the sequence is defined recursively with an initial term given as \(a_1 = 256\) and the subsequent terms defined by \(a_{n+1} = \sqrt{a_n} \).
02
Calculate the first term
The first term is given directly: \(a_1 = 256\).
03
Calculate the second term
Using \(a_{n+1} = \sqrt{a_n}\), find the second term: \(a_2 = \sqrt{256} = 16\).
04
Calculate the third term
Next, use the same formula to find the third term: \(a_3 = \sqrt{16} = 4\).
05
Calculate the fourth term
Again, apply the same formula to find the fourth term: \(a_4 = \sqrt{4} = 2\).
06
List the first four terms
The first four terms of the sequence are: 256, 16, 4, and 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
sequence terms
In mathematics, a sequence is an ordered list of numbers. Each number in the sequence is called a term. To understand sequences, it helps to break them down into their individual components. In this exercise, we deal with a special type of sequence known as a recursively defined sequence. Here, each term depends on the previous term. Let's dive into this key concept further. Sequences can be finite or infinite, and terms are typically denoted by a letter with a subscript to indicate their position in the sequence. For instance, the first term is usually denoted by \(a_1\), the second term by \(a_2\), and so on.
square roots
Square roots are an essential mathematical operation frequently used in sequences involving recursive definitions. The square root of a number \(x\) is a value \(y\) such that \(y^2 = x\). It means that \(y\) multiplied by itself equals \(x\). This operation is denoted by the radical symbol \( \,\sqrt{}\, \). In our recursive sequence, each term is derived by taking the square root of the previous term. For example, if \(a_1 = 256\), then the next term \(a_{n+1}\) is \( \,\sqrt{256} = 16 \, \). This process continues with each term: \( \,\sqrt{16} = 4 \, \) and \( \,\sqrt{4} = 2 \, \). It's important to understand the basic properties of square roots:
- \( \,\sqrt{a^2} = a \, \)
- \( \,\sqrt{ab} = \sqrt{a} \cdot \sqrt{b} \, \)
- \( \,\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \, \)
initial term
The initial term is the starting point of a sequence. In recursively defined sequences, it is crucial as it determines all subsequent terms. Given the initial term \(a_1\), each next term is calculated based on this value. For example, in the given exercise, our initial term is \(a_1 = 256\). This pre-set value is the anchor of our sequence. From this initial term, we apply a formula repeatedly to get the next terms. Understanding and identifying the initial term correctly ensures you can correctly generate and understand the sequence. For example,
- The first term \(a_1 = 256\)
- The second term \(a_2 = \sqrt{256} = 16\)
- The third term \(a_3 = \sqrt{16} = 4\)
- The fourth term \(a_4 = \sqrt{4} = 2\)