Chapter 11: Problem 67
Find a formula for the \(n\) th term of the sequence \(\sqrt{2}, \quad \sqrt{2 \sqrt{2}}, \quad \sqrt{2 \sqrt{2 \sqrt{2}}}, \quad \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}, \ldots\)
Short Answer
Expert verified
The formula for the \( n \)th term is \( 2^{1 - \frac{1}{2^n}} \).
Step by step solution
01
Identify the Pattern
Observe that each term in the sequence has a nested radical structure. The first term is \( \sqrt{2} \), the second term is \( \sqrt{2 \times \sqrt{2}} \), and the third term is \( \sqrt{2 \times \sqrt{2 \times \sqrt{2}}} \). Notice that this pattern can be expressed in terms of powers of 2.
02
Express Radicals in Exponential Form
Rewrite the terms as exponents. The first term \( \sqrt{2} \) can be rewritten as \( (2^{1/2}) \). The second term \( \sqrt{2 \times \sqrt{2}} \) is equivalent to \( (2^{1/2})^{3/2} = 2^{3/4} \). Observe that each subsequent term builds upon the exponent from the previous term.
03
Establish a Recurrence for Exponents
Define the exponent of \( 2 \) in the \( n \)th term as \( a_n \). Based on the pattern, we see \( a_1 = \frac{1}{2}, a_2 = \frac{3}{4} \), and \( a_3 = \frac{7}{8} \). Notice how \( a_n = 1 - \frac{1}{2^n} \).
04
Derive the General Formula
Recognizing that the sequence follows \( a_n = 1 - \frac{1}{2^n} \), the \( n \)th term of the sequence is given by \( \sqrt[n]{2} = 2^{1 - \frac{1}{2^n}} \). This is equivalent to \( 2^{1 - \frac{1}{2^n}} = 2^{1} \times 2^{- \frac{1}{2^n}} \).
05
Formulate the nth Term
The \( n \)th term of the sequence can be written as \( 2^{1 - \frac{1}{2^n}} \), which represents the nested radical expression in its exponential form.Therefore, the formula for the \( n \)th term of the sequence is \( 2^{1 - \frac{1}{2^n}} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Form
The concept of exponential form is a powerful tool in mathematics, helping to simplify expressions with repeated multiplication. By using bases and exponents, complex calculations can be expressed in a more compact form. In our sequence, we transform endless products and square roots into simpler exponential terms. For instance:
- The number \( \sqrt{2} \) can be rewritten as \( 2^{1/2} \).
- The expression \( \sqrt{2 \times \sqrt{2}} \) becomes \( 2^{3/4} \) when expressed exponentially.
Nested Radicals
Nested radicals are expressions containing a square root within another square root. These can often appear daunting and complex. In the given sequence:
- The first term has a simple radical: \( \sqrt{2} \).
- The second term adds another layer: \( \sqrt{2 \times \sqrt{2}} \).
- Each additional term introduces further nesting, like \( \sqrt{2 \times \sqrt{2 \times \sqrt{2}}} \).
Recurrence Relation
Recurrence relations define sequences in a way that each term is based on the preceding ones. This method of expression is particularly useful for sequences where a simple pattern is detectable, as in our exercise. With the sequence provided, the exponents of 2 form a recurrence relation:
- The first exponent is \( a_1 = rac{1}{2} \).
- The second exponent is \( a_2 = rac{3}{4} \).
- The relation is modeled as \( a_n = 1 - rac{1}{2^n} \).