Problem 22 - Precalculus with Limits

Question

In Exercises 9-32, write the first five terms of the sequence. (Assume that \( n \) begins with 1.) \( a_n = \dfrac{2^n}{3^n} \)

Step-by-Step Solution

Answer

The first five terms of the sequence are \(\frac{2}{3}\), \(\frac{4}{9}\), \(\frac{8}{27}\), \(\frac{16}{81}\), \(\frac{32}{243}\)

Step 1: Understand the rule for the sequence

The rule given for the sequence is \(a_n = \dfrac{2^n}{3^n}\). This means that to find any term \(a_n\) in the sequence, you raise 2 to the power \(n\) and divide the result by 3 to the power \(n\). Now apply this rule for \(n\) from 1 to 5 to get the first five terms of the sequence.

Step 2: Calculate \(a_1\)

Substitute \(n = 1\) into the sequence rule to find \(a_1\): \(a_1 = \dfrac{2^1}{3^1} = \dfrac{2}{3}\).

Step 3: Calculate \(a_2\)

Substitute \(n = 2\) into the sequence rule to find \(a_2\): \(a_2 = \dfrac{2^2}{3^2} = \dfrac{4}{9}\).

Step 4: Calculate \(a_3\)

Substitute \(n = 3\) into the sequence rule to find \(a_3\): \(a_3 = \dfrac{2^3}{3^3} = \dfrac{8}{27}\).

Step 5: Calculate \(a_4\)

Substitute \(n = 4\) into the sequence rule to find \(a_4\): \(a_4 = \dfrac{2^4}{3^4} = \dfrac{16}{81}\).

Step 6: Calculate \(a_5\)

Substitute \(n = 5\) into the sequence rule to find \(a_5\): \(a_5 = \dfrac{2^5}{3^5} = \dfrac{32}{243}\).

Key Concepts

Geometric Sequences

A geometric sequence is a series of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. For example, in the sequence 2, 4, 8, 16, ..., each term is obtained by multiplying the previous term by the common ratio of 2.

In the given exercise, the sequence is defined by the formula \( a_n = \frac{2^n}{3^n} \) is a geometric sequence with a common ratio of \( \frac{2}{3} \). Each term decreases in size as n increases because the ratio is less than 1. This behavior is characteristic of a geometric sequence with a common ratio between -1 and 1, indicating a sequence that converges toward zero as n approaches infinity.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. In the context of sequences, the general form of an exponential function is \( b^n \) where \( b \) is the base, and \( n \) is the exponent, which is often the term number in the sequence.

The sequence rule \( a_n = \frac{2^n}{3^n} \) derives from exponential functions. The numerator \( 2^n \) represents an exponential function with base 2, and the denominator \( 3^n \) with base 3. When the base is larger than 1, the function's values increase exponentially as n grows. Conversely, values decrease exponentially when the base is a fraction, as is the case with the given sequence.

Sequence Terms Calculation

Sequence terms calculation involves finding the specific terms of a sequence using a defined rule. In a geometric sequence, any term \(a_n\) can be calculated using the formula \(a_n = a_1 \cdot r^{(n-1)}\), where \(a_1\) is the first term and \(r\) is the common ratio.

For the exercise at hand, \(a_n\) is found by substituting the term position \(n\) into the sequence rule \(\frac{2^n}{3^n}\). Since exponential calculations can be complex, careful manipulation of powers and exponents is necessary—such as expressing the sequence in terms of a single power to simplify the calculations, as \(\left(\frac{2}{3}\right)^n\).

Powers and Exponents

Powers and exponents are fundamental concepts in mathematics that represent repeated multiplication. A number raised to an exponent, or power, indicates how many times to multiply that number by itself. For instance, \( 2^3 = 2 \times 2 \times 2 = 8 \).

In our sequence exercise, \( 2^n \) and \( 3^n \) are representations of 2 and 3 raised to the power of n, respectively. Dividing these expressions exhibits the properties of exponents, particularly \( \frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n \), which simplifies the computation of our sequence's terms.