Problem 69 - Algebra and Trigonometry

Question

Find a formula for the nth 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 of the sequence is \( 2^{(1 - \frac{1}{2^n})} \).

Step 1: Identify the First Term

The sequence starts with \( \sqrt{2} \). We recognize that \( \sqrt{2} \) can be written as \( 2^{1/2} \). Therefore, the first term, \( a_1 \), is \( 2^{1/2} \).

Step 2: Interpret the Second Term

The second term is \( \sqrt{2 \times 2^{1/2}} \). This can be written as \( (2 \times 2^{1/2})^{1/2} = (2^{1+1/2})^{1/2} = 2^{(1+1/2) \times 1/2} = 2^{3/4} \). Thus, the second term, \( a_2 \), is \( 2^{3/4} \).

Step 3: Derive the Third Term

Proceed similarly with the third term: it is \( \sqrt{2 \times 2^{3/4}} \). We calculate \( (2 \times 2^{3/4})^{1/2} = (2^{1+3/4})^{1/2} = 2^{(1+3/4) \times 1/2} = 2^{7/8} \). So, the third term, \( a_3 \), is \( 2^{7/8} \).

Step 4: Explore the Pattern

Observe the powers: 1/2, 3/4, 7/8. Each exponent of 2 for term \( a_n \) appears to follow the pattern of \( a_n = 2^{(1 - \frac{1}{2^n})} \). For instance, the third exponent 7/8 is computed as \( 1 - 1/8 \).

Step 5: Formulate the Nth Term

Recognize that each term in the sequence can be described by the formula \( a_n = 2^{(1 - \frac{1}{2^n})} \). This formula captures the observed pattern in the exponents as each subsequent term multiplies the inner square root.

Key Concepts

Exponents

In mathematics, exponents are used to represent repeated multiplication of a number by itself. For example, \( 2^3 \) means \( 2 \times 2 \times 2 \), which equals 8. An exponent simply tells us how many times a base number is used as a factor.
Exponents make it easier to write and work with large numbers. Instead of writing a long series of multiplications, we can simplify it using exponents.
In the exercise sequence, each term is expressed using exponents. For example, \( \sqrt{2} \) is represented as \( 2^{1/2} \). Understanding how to manipulate exponents is crucial for interpreting this sequence correctly. As seen in the step by step solution, exponents help us find patterns and formulate rules for sequences.

Square Roots

A square root of a number is a value that, when multiplied by itself, gives the original number. The square root of 4 is 2, because \( 2 \times 2 = 4 \). Square roots are represented by the radical symbol \( \sqrt{} \).
In sequences involving square roots, each term is an iteration of nested square roots. In the given sequence, the first term is \( \sqrt{2} \), which is an immediate indication to change into a form that can help identify a pattern. By expressing square roots as exponents, like \( \sqrt{2} = 2^{1/2} \), it enables us to quantitatively track changes from one term to the next.

• This transformation simplifies recognizing the exponential growth pattern of the sequence.
• It also assists in applying the patterns seen in conversions and helps weight each subsequent term in the sequence.

Patterns in Sequences

Patterns in sequences are fundamental in mathematics, offering a systematic way to predict future terms. Recognizing these patterns simplifies complex sequences and assists in finding the nth term.
In the provided sequence, the pattern emerges within the exponents. Starting from \( 2^{1/2} \) and proceeding to \( 2^{3/4} \) and \( 2^{7/8} \), we observe an increment in the numerator of the exponents.

• The denominators double (2 becomes 4, then 8) while the numerators follow the pattern of adding consecutive natural numbers \( 1, 2, 3... \).
• This pattern is crucial as it allows us to deduce the general formula, capturing the change as \( 1 - \frac{1}{2^n} \) for the nth term.

Identifying such patterns is key to understanding sequences better. It becomes a powerful tool for exploring how repetitions and recursive relationships can be formulated in algebraic terms.

Nth Term Formula

The nth term formula is a mathematical expression that allows us to find any term in a sequence without generating all previous terms. It encapsulates the pattern discovered into a single, easy-to-use equation.
In our sequence, the nth term is deduced to be \( 2^{(1 - \frac{1}{2^n})} \).

• This formula elegantly captures the evolving pattern of exponents as each term is calculated by manipulating the initial square root transformation into exponential growth.
• It succinctly encapsulates all necessary operations within a consistent rule that applies to any part of the sequence.

Understanding and creating such formulas is beneficial because:

• It saves time, as there is no need to recalculate each term manually.
• It provides a deeper understanding of the sequence's structure and its behavior over a large number of terms.

Mastering the skill of deriving the nth term formula is a fundamental mathematical exercise, revealing the elegance of patterns and the power of algebraic representation.