Problem 24 - Thomas Calculus

Question

In Exercises \(13-26,\) find a formula for the \(n\) th term of the sequence. $$ \frac{1}{25}, \frac{8}{125}, \frac{27}{625}, \frac{64}{3125}, \frac{125}{15,625}, \ldots $$

Step-by-Step Solution

Answer

The \(n\)th term of the sequence is \(\frac{n^3}{5^{n+1}}\).

Step 1: Observe the Pattern in the Numerators

The numerators of the sequence are as follows: 1, 8, 27, 64, 125. Notice that these numbers correspond to perfect cubes: \(1 = 1^3\), \(8 = 2^3\), \(27 = 3^3\), \(64 = 4^3\), and \(125 = 5^3\). Hence, we identify that the numerator for the \(n\)th term is \(n^3\).

Step 2: Observe the Pattern in the Denominators

The denominators of the sequence are 25, 125, 625, 3125, 15625. Recognize that these can be expressed as powers of 5: \(25 = 5^2\), \(125 = 5^3\), \(625 = 5^4\), \(3125 = 5^5\), and \(15625 = 5^6\). Thus, the denominator for the \(n\)th term is given by \(5^{n+1}\).

Step 3: Combine the Patterns to Form the General Term

From the patterns observed, the general form for the \(n\)th term of the sequence can be constructed. Using the numerators and denominators obtained, the \(n\)th term is \(\frac{n^3}{5^{n+1}}\).

Step 4: Validate the Formula with Initial Terms

To ensure correctness, substitute initial values of \(n\) into \(\frac{n^3}{5^{n+1}}\). For \(n=1\), \(\frac{1^3}{5^{1+1}} = \frac{1}{25}\), for \(n=2\), \(\frac{2^3}{5^{2+1}} = \frac{8}{125}\), which match the given sequence values. Continue this for additional terms to confirm the pattern holds.

Key Concepts

Numerator Patterns

When examining sequences, the pattern of the numerators is often the first clue that can help us identify the sequence's structure. In our given sequence, the numerators are 1, 8, 27, 64, and 125. Recognizing patterns is key, and here these numbers align with perfect cubes. Specifically, we have:

• 1 as the cube of 1: \(1^3 = 1\)
• 8 as the cube of 2: \(2^3 = 8\)
• 27 as the cube of 3: \(3^3 = 27\)
• 64 as the cube of 4: \(4^3 = 64\)
• 125 as the cube of 5: \(5^3 = 125\)

The pattern indicates that the numerator for each term is the cube of the term's position in the sequence. This means for the \(n\)th term of the sequence, the numerator can be expressed as \(n^3\). This cube pattern is a frequent occurrence in mathematical sequences, making it an essential pattern to identify for simplifying sequence analysis.

Denominator Patterns

The denominators in a sequence can also reveal a noticeable pattern, as is evident in our sequence. The denominators provided are 25, 125, 625, 3125, and 15625. These numbers are powers of 5:

• \(25\) is \(5^2\)
• \(125\) is \(5^3\)
• \(625\) is \(5^4\)
• \(3125\) is \(5^5\)
• \(15625\) is \(5^6\)

We can see that the pattern involves increasing the exponent of 5 by 1 for each subsequent term, starting from 2. Therefore, the denominator for the sequence's \(n\)th term can be described by the formula \(5^{n+1}\). Recognizing powers of a base number like this is a common technique in sequences to establish a general rule for the terms.

General Term of a Sequence

After identifying patterns in the numerators and denominators, combining these can help in forming the general term of the sequence. From our observations:

• The numerator for the \(n\)th term is \(n^3\).
• The denominator for the \(n\)th term is \(5^{n+1}\).

A sequence's general term provides a formula that allows us to calculate any term directly without listing all preceding terms. Together, these observations yield the general formula for the \(n\)th term as:\[\frac{n^3}{5^{n+1}}\]To ensure the formula's accuracy, it is beneficial to validate it with the initial terms of the sequence. For instance, when \(n=1\), substituting into the general term gives \(\frac{1^3}{5^{1+1}} = \frac{1}{25}\). Similarly, for \(n=2\), substituting gives \(\frac{2^3}{5^{2+1}} = \frac{8}{125}\). These results match the original sequence accurately, confirming the validity of the formula. Understanding and deriving the general term is vital for analyzing and predicting sequence behavior in mathematics.