Problem 32 - Algebra and Trigonometry

Question

\(25-32\) . Find the \(n\) th term of a sequence whose first several terms are given. $$ 1, \frac{1}{2}, 3, \frac{1}{4}, 5, \frac{1}{6}, \dots $$

Step-by-Step Solution

Answer

The nth term is \(2k - 1\) if n is odd, and \(\frac{1}{2k}\) if n is even, where \(k = \frac{n+1}{2}\) if n is odd and \(k = \frac{n}{2}\) if n is even.

Step 1: Identify the Pattern

Observe the sequence: the terms alternate between whole numbers and fractions. The pattern for whole numbers is 1, 3, 5, ... which are odd numbers starting from 1. For the fractions, the pattern is \( \frac{1}{2}, \frac{1}{4}, \frac{1}{6}, ... \) which are inverses of even numbers.

Step 2: Determine Subsequence Formulas

For the whole numbers, they follow the sequence: 1, 3, 5, which is the arithmetic sequence given by the formula \( a_n = 2n - 1 \). For the fractions, starting with \( \frac{1}{2} \), they are in the sequence \( \frac{1}{2n} \), where n is an integer starting from 1.

Step 3: Define the Sequence Structure

Note that for odd-numbered positions in the sequence, the terms are odd numbers, and for even-numbered positions, the terms are fractions. Thus, terms at odd positions \( 2k-1 \) use formula \( 2k - 1 \) and terms at even positions \( 2k \) use \( \frac{1}{2k} \).

Step 4: Formulate the General Term

The nth term of the sequence is determined by whether n is odd or even. If n is odd \((n=2k-1)\), the term is \(2k - 1\). If n is even \((n=2k)\), the term is \(\frac{1}{2k}\).

Key Concepts

Arithmetic Sequence

Sequences are fascinating, and an arithmetic sequence is one of the simplest. In an arithmetic sequence, the difference between consecutive terms is constant. This difference is known as the "common difference." Consider the sequence of numbers such as 3, 5, 7, 9, and so on. Each number is obtained by adding 2, which is the common difference, to the previous number. In our example, starting with odd numbers 1, 3, 5..., these numbers also form an arithmetic sequence with a common difference of 2. You can express this in a formula: - Each term can be found by adding the common difference, 2, to the previous term.Knowing the formula for an arithmetic sequence helps you predict future terms without writing out all the numbers. So, if you want to find a specific future term, you use this general formula: if the first term is 'a' and the common difference is 'd', then the n-th term is given by: \[ a_n = a + (n-1)d \]Remembering this formula can save you time when working with arithmetic sequences.

n-th Term

Many students find calculating the n-th term of a sequence daunting, but it's easier than it seems. The n-th term provides a way to find any term in a sequence without listing all previous numbers. This is crucial in math because it allows you to directly calculate any term you need.In the exercise provided, terms alternate between two types, requiring different formulas depending on whether the position is odd or even:- For odd positions, use the formula for finding an odd number in its sequence: \( 2k - 1 \).- For even positions, use the formula for the fractions: \( \frac{1}{2k} \).Notice that 'k' helps determine the position in each subset of the sequence. Understanding whether 'n' is odd or even helps you choose the correct formula, ensuring you solve for the correct n-th term efficiently.

Subsequence Formulas

Sometimes a sequence may contain hidden subsequences, as in our exercise. A subsequence is simply a sequence extracted from another by omitting some elements without changing the order.In this case, notice two separate patterns:

• The odd-numbered sequence: uses the arithmetic sequence \( a_n = 2n - 1 \) for terms like 1, 3, 5, etc.
• The fraction sequence: follows formula \( \frac{1}{2n} \) for terms like \( \frac{1}{2}, \frac{1}{4}, \frac{1}{6} \), based on even-numbered positions.

Recognizing a subsequence lets you apply the right formula and explicitly understand the overall pattern. Each subsequence provides a clear method to find elements specific to parts of a sequence, aiding in complex problems.