Question

The given pattern continues. Write down the nth term of a sequence \(\left\\{a_{n}\right\\}\) suggested by the pattern. \(\frac{2}{3}, \frac{4}{9}, \frac{8}{27}, \frac{16}{81}, \ldots\)

Short answer

\( \left( \frac{2}{3} \right)^n \)

Step-by-step solution

- Observe the Given Pattern

Look at the sequence: \( \frac{2}{3}, \frac{4}{9}, \frac{8}{27}, \frac{16}{81}, \ldots \). Notice how the numerator and denominator change.

- Identify the Numerator Pattern

Observe the numerators: 2, 4, 8, 16, ... Each numerator is a power of 2. Specifically, the numerator is given by \( 2^n \).

- Identify the Denominator Pattern

Observe the denominators: 3, 9, 27, 81, ... Each denominator is a power of 3. Specifically, the denominator is given by \( 3^n \).

- Formulate the nth Term

Combine the patterns identified for the numerator and the denominator. The nth term \( a_n \) of the sequence is given by \( a_n = \frac{2^n}{3^n} \).

- Simplify the nth Term

Simplify \( \frac{2^n}{3^n} \) to \( \left( \frac{2}{3} \right)^n \). So, the nth term is \( \left( \frac{2}{3} \right)^n \).

Key concepts

nth term formula

In sequences, the nth term formula is crucial. It gives a general expression to calculate any term in the sequence without listing all prior terms. For example, if you know the nth term, you can find the 100th term without finding the 99 terms before it.
For the sequence given in the exercise: \( \frac{2}{3}, \frac{4}{9}, \frac{8}{27}, \frac{16}{81} \), the nth term is identified by analyzing patterns in the numerators and denominators.
We determined that both follow specific exponential patterns: the numerator as \( 2^n \) and the denominator as \( 3^n \). Combining these, the nth term is \( a_n = \frac{2^n}{3^n} \). This can be further simplified to \( \bigg( \frac{2}{3} \bigg)^n \) for simplicity.

pattern recognition

Recognizing patterns is the key to solving sequence problems. Look at repeated changes or consistent rules.
In the given sequence \( \frac{2}{3}, \frac{4}{9}, \frac{8}{27}, \frac{16}{81} \), the numerators and denominators change consistently, making it easier to spot the pattern.
The numerators double with each term: 2, 4, 8, 16... This is a clear sign they are powers of 2, i.e., \( 2^1, 2^2, 2^3, 2^4 \). Similarly, the denominators are 3, 9, 27, 81, which are powers of 3, i.e., \( 3^1, 3^2, 3^3, 3^4 \). Recognizing this pattern allows you to generalize and form the nth term formula for the sequence.
Mastering pattern recognition enables you to swiftly identify and formulate sequences in algebra.

powers of integers

The concept of powers of integers is fundamental in algebra, especially in sequences. Powers of integers involve raising a base number to an exponent. For example, \( 2^3 = 2 \times 2 \times 2 = 8 \) and \( 3^2 = 3 \times 3 = 9 \).
In the provided sequence, the numerators are integer powers of 2: \( 2^1, 2^2, 2^3, 2^4 \) which are 2, 4, 8, 16 respectively. The denominators follow a similar pattern but with base 3: \( 3^1, 3^2, 3^3, 3^4 \) which are 3, 9, 27, 81 accordingly.
Understanding and identifying these powers aids in simplifying complex sequences. In our sequence, because both the numerator and the denominator are powers of integers, we could simplify the nth term to \( \bigg( \frac{2}{3} \bigg)^n \), showcasing how powers operate within fractions as well.