Problem 69 - Precalculus Mathematics for Calculus

Question

Find a formula for the \(n\) th term of the sequence \(\sqrt{2}, \quad \sqrt{2 \sqrt{2}}, \quad \sqrt{2 \sqrt{2 \sqrt{2}}}, \quad \sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}, \ldots\) [Hint: Write each term as a power of \(2 .]\)

Step-by-Step Solution

Answer

The nth term is \(2^{1 - \frac{1}{2^n}}\).

Step 1: Analyze the Sequence

Observe the given sequence. Each term involves nested square roots. The terms are:\( \sqrt{2}, \quad \sqrt{2\sqrt{2}}, \quad \sqrt{2\sqrt{2\sqrt{2}}}, \quad \sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}} \).

Step 2: Express Terms in Simplified Form

Write each of the terms as a power of 2. For the first term, \( \sqrt{2} = 2^{1/2} \). The second term is \( \sqrt{2\sqrt{2}} = 2^{1/2} \cdot 2^{1/4} = 2^{3/4} \). Similarly, the third term \( \sqrt{2\sqrt{2\sqrt{2}}} = 2^{3/4} \cdot 2^{1/8} = 2^{7/8}\).

Step 3: Identify the Pattern

Observe the exponents in each term: The first term has an exponent 1/2, the second 3/4, and so on. Notice that with each term, we add a power of \(1/2^n\) to the previous exponent.

Step 4: Derive a General Formula

From Step 3, the exponent of each term follows the pattern: \(1/2, \quad 3/4 = 1/2 + 1/4, \quad 7/8 = 3/4 + 1/8 = 1/2 + 1/4 + 1/8, \ldots\). Thus, for the nth term, the exponent is the sum of a geometric series: \(\sum_{k=1}^{n} \frac{1}{2^k}\).

Step 5: Evaluate the Sum of the Series

The series \(\sum_{k=1}^{n} \frac{1}{2^k}\) is a geometric series with the first term \(a = \frac{1}{2}\) and common ratio \(r = \frac{1}{2}\). The sum of the first \(n\) terms is given by \(S_n = \frac{a(1-r^n)}{1-r} = \frac{\frac{1}{2}(1-\frac{1}{2^n})}{\frac{1}{2}} = 1 - \frac{1}{2^n}\).

Step 6: Formulate the nth Term

The nth term of the sequence is such that the base is 2 and the overall exponent from Step 5 is \(1 - \frac{1}{2^n}\). Thus, the nth term is \(2^{1 - \frac{1}{2^n}}\).

Key Concepts

nth term

In a sequence, finding the nth term means we want to derive a general formula that lets us calculate any term in the sequence as needed. In this specific exercise, we dealt with nested square roots forming a pattern. Each term could be expressed as a power of 2. As identified in the solution, the exponent of each term forms a pattern which is essentially the sum of a geometric series. For the nth term, we derive the formula:

• First term: 2 raised to the power of \( 1/2 \)
• Second term: 2 raised to the power of \( 3/4 \)
• Third term: 2 raised to the power of \( 7/8 \)

The pattern continues, and thus the nth term is expressed as \( 2^{1 - \frac{1}{2^n}} \). This formula enables us to compute any term without repeating all steps.

geometric series

A geometric series is a series of terms where each term after the first is the product of the previous term and a fixed, non-zero number called the common ratio. In this exercise, the exponents of 2 in the sequence form a geometric series:

• First term exponent: \( \frac{1}{2} \)
• Second term exponent: \( \frac{1}{2} + \frac{1}{4} \)
• Third term exponent: \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \)

The common ratio here is \( \frac{1}{2} \). The formula for the sum of the first n terms of a geometric series is given by:\[S_n = \frac{a(1-r^n)}{1-r}\]Where \( a \) is the first term, and \( r \) is the common ratio. Applying this, we find that the sum of the exponents for the nth term is \( 1 - \frac{1}{2^n} \).

exponents

Exponents are vital in mathematics as they represent repeated multiplication. In this exercise, each term of the sequence can be viewed as 2 raised to a certain power (the exponent).When multiplying numbers with the same base, we add the exponents:\[ a^m \times a^n = a^{m+n} \]Applying this to our sequence, the terms can be rewritten as:

• First term: \( \sqrt{2} = 2^{1/2} \)
• Second term: \( 2^{1/2} \times 2^{1/4} = 2^{3/4} \)
• Third term: \( 2^{3/4} \times 2^{1/8} = 2^{7/8} \)

This consistent pattern in exponents allows us to derive the nth term's formula by summing up the series of fractional exponents.

square roots

Square roots are a fundamental concept. The square root of a number is a value that, when squared, gives the original number. For instance, \( \sqrt{2} \) is a number which, when multiplied by itself, equals 2.In this exercise, each term starts with square roots that get increasingly nested. The relationship between square roots and exponents can be established as:

• \( \sqrt{x} = x^{1/2} \)
• Nested square roots like \( \sqrt{2\sqrt{2}} = (2^{1/4} \times 2^{1/2}) = 2^{3/4} \)

Understanding how square roots translate into fractional exponents, help us to express each sequence term in a clean exponential form. This insight simplifies these calculations and aids in finding the nth term formula: \( 2^{1 - \frac{1}{2^n}} \).