Problem 55 - Precalculus with Limits

Question

In Exercises 47-62, write an expression for the apparent \(n\)th term of the sequence. (Assume that \( n \) begins with 1.) \( 1, \dfrac{1}{4}, \dfrac{1}{9}, \dfrac{1}{16}, \dfrac{1}{25}, \dots \)

Step-by-Step Solution

Answer

The apparent \(n\)th term of the sequence is \(\dfrac{1}{n^2}\).

Step 1: Observation of the sequence

The first step is to observe the sequence - the key feature to note is each term is 1 divided by a perfect square number. Additionally, the perfect square number is the square of the index of that term in the sequence. For example, the first term is \(\dfrac{1}{1}\), the second term is \(\dfrac{1}{4}\) which is \(\dfrac{1}{2^2}\), the third term is \(\dfrac{1}{9}\) which is \(\dfrac{1}{3^2}\), and so on. This suggests that the appearance of the \(n\)th term follow a rule related to \(n\).

Step 2: Formulate the expression

Based on the observation, we can see that each term in this sequence is the reciprocal of the square of each respective term's index. This leads to the formulation that the \(n\)th term of this sequence can be calculated as \(\dfrac{1}{n^2}\). This is because for any \(n\), the value at that point in the sequence will be \(\dfrac{1}{n^2}\).

Key Concepts

Arithmetic Series

An arithmetic series is a particular type of sequence where each term is derived by adding a fixed number, called the common difference, to the previous term. For instance, in the sequence 2, 5, 8, 11,... the common difference is 3, because each subsequent term is obtained by adding 3 to the previous one.

In contrast, the exercise provided does not form an arithmetic series because the terms are not obtained by adding a constant. Instead, the sequence involves division by increasingly large perfect squares, which alters the pattern entirely.

Arithmetic series are useful in calculations because they allow for easy prediction and summation with formulas. Although not applicable in the current exercise, understanding that this sequence behaves differently can help prevent confusion in future problems.

N-th Term

The "n-th term" of a sequence provides a way to find any term in the sequence without listing all previous terms. It is an expression that describes the value at any position "n" in a sequence.

In the given sequence, the terms are 1, \(\frac{1}{4}\), \(\frac{1}{9}\), \(\frac{1}{16}\), \(\frac{1}{25}\), ..., and each term can be described using the formula for the "n-th term". By using this expression, \(a_n = \frac{1}{n^2}\), you can find the value at position \(n\) by simply substituting \(n\).

This approach saves time and reduces errors, especially when dealing with very large or complex sequences. Recognizing patterns and deriving the formula for the "n-th term" are essential skills in sequence analysis.

Recursive Sequence

A recursive sequence is a sequence in which each term is defined based on previous terms. Unlike a sequence with a defined "n-th term" formula, recursive sequences rely on the value of preceding terms for calculation.

For example, a simple recursive sequence might be \(b_1 = 1\) and \(b_{n+1} = b_n + 3\), where each term is the previous term plus three. However, our sequence \(1, \frac{1}{4}, \frac{1}{9}, \frac{1}{16}, \frac{1}{25},...\) does not follow recursion.

Instead, it's defined explicitly through the formula \(a_n = \frac{1}{n^2}\), which is why it cannot directly be expressed recursively. Recognizing the difference between recursive and explicit ("n-th term") sequences is crucial for solving various sequence-related problems.