Problem 56 - College Algebra

Question

At which term does the sequence \(\left\\{\frac{1}{2187}, \frac{1}{729}, \frac{1}{243}, \frac{1}{81} \ldots\right\\}\) begin to have integer values?

Step-by-Step Solution

Answer

The sequence begins to have integer values at the 8th term.

Step 1: Determine the Pattern of the Sequence

The sequence is given as \( \left\{ \frac{1}{2187}, \frac{1}{729}, \frac{1}{243}, \frac{1}{81}, \ldots \right\} \). Observe how each term relates to the previous one. Calculating the ratio between consecutive terms: \(\frac{1}{729} \div \frac{1}{2187} = 3\), \(\frac{1}{243} \div \frac{1}{729} = 3\), and \(\frac{1}{81} \div \frac{1}{243} = 3\). This suggests that the sequence is geometric with common ratio 3.

Step 2: Write the General Term Formula

To find the general formula of the sequence, recall that for a geometric sequence, the general term \(a_n\) is given by \(a_n = a_1 \cdot r^{n-1}\), where \(a_1\) is the first term, and \(r\) is the common ratio. Here, \(a_1 = \frac{1}{2187}\) and \(r = 3\), so \(a_n = \frac{1}{2187} \cdot 3^{n-1}\).

Step 3: Determine When the Sequence Yields Integers

We want to find \(n\) such that \(a_n\) is an integer. Set \(a_n = \frac{1}{2187} \cdot 3^{n-1}= k\), where \(k\) is an integer. Simplifying \(3^{n-1}\) needs to cancel out the denominator 2187. Note that \(2187 = 3^7\), so \(3^{n-1}\cdot \frac{1}{3^7}\) is an integer when \(n-1 \geq 7\).

Step 4: Solve for n

From \(n-1 \geq 7\), we find \(n \geq 8\). Therefore, the smallest integer \(n\) for which the sequence term is an integer is \(n = 8\).

Key Concepts

Common Ratio

In a geometric sequence, the "common ratio" is a pivotal concept. This is the constant factor between consecutive terms of the sequence. To find this ratio, divide any term by the previous term. For the sequence given in our problem:

• \( \frac{1}{729} \div \frac{1}{2187} = 3 \)
• \( \frac{1}{243} \div \frac{1}{729} = 3 \)
• \( \frac{1}{81} \div \frac{1}{243} = 3 \)

You will notice that the common ratio is 3 throughout. This means every term is 3 times larger than the one before it. In this specific problem, understanding this constant multiplier helps identify patterns and predict future terms.

General Term Formula

The general term formula of a geometric sequence allows us to find any term in the sequence. It is expressed as: \[ a_n = a_1 \cdot r^{n-1} \]Where:- \(a_n\) is the term you want to find,- \(a_1\) is the first term of the sequence,- \(r\) is the common ratio, and- \(n\) is the term number.In our sequence, the first term \(a_1\) is \(\frac{1}{2187}\), and the common ratio \(r\) is 3.Thus, the general term formula is:\[ a_n = \frac{1}{2187} \cdot 3^{n-1} \]This formula is very useful as it allows for easy calculation of any term in the sequence without needing to calculate all the preceding terms.

Integer Sequence

An integer sequence is where every term is a whole number. Our task is to find when terms in this geometric sequence become whole numbers. This depends on how the general term formula aligns with integers. For the sequence \( \frac{1}{2187} \cdot 3^{n-1} \), simplifying that to an integer requires \[ \frac{3^{n-1}}{2187} \]Considering the denominator, note that \(2187 = 3^7\). So for the general term to be an integer,\( 3^{n-1} \) should be greater than or equal to \( 3^7 \), i.e., \( n-1 \geq 7 \). Solving this inequality gives \( n \geq 8 \).Thus, the sequence begins to yield integer values at the 8th term, because it first matches or exceeds the power of the denominator. Understanding when and why a sequence becomes an integer is crucial for pattern prediction and real-world applications.