Problem 19 - Precalculus: Mathematical for Calculus

Question

Find the first five terms of the sequence and determine if it is arithmetic. If it is arithmetic, find the common difference and express the \(n\) th term of the sequence in the standard form $$a_{n}=a+(n-1) d$$ $$a_{n}=\frac{1}{1+2 n}$$

Step-by-Step Solution

Answer

The sequence is not arithmetic. The differences between terms are not constant.

Step 1: Identify the Sequence Formula

We are given the sequence defined by \(a_n = \frac{1}{1+2n}\). We need to find the first five terms by substituting different values of \(n\).

Step 2: Calculate the First Five Terms

To find the first term, substitute \(n = 1\) into the formula: \(a_1 = \frac{1}{1+2(1)} = \frac{1}{3}\).For the second term, substitute \(n = 2\): \(a_2 = \frac{1}{1+2(2)} = \frac{1}{5}\).For the third term, substitute \(n = 3\): \(a_3 = \frac{1}{1+2(3)} = \frac{1}{7}\).For the fourth term, substitute \(n = 4\): \(a_4 = \frac{1}{1+2(4)} = \frac{1}{9}\).Finally, for the fifth term, substitute \(n = 5\): \(a_5 = \frac{1}{1+2(5)} = \frac{1}{11}\).

Step 3: Determine if the Sequence is Arithmetic

An arithmetic sequence has a constant difference between consecutive terms. Calculate the differences: 1. \(a_2 - a_1 = \frac{1}{5} - \frac{1}{3} = -\frac{2}{15}\).2. \(a_3 - a_2 = \frac{1}{7} - \frac{1}{5} = -\frac{2}{35}\).3. \(a_4 - a_3 = \frac{1}{9} - \frac{1}{7} = -\frac{2}{63}\).4. \(a_5 - a_4 = \frac{1}{11} - \frac{1}{9} = -\frac{2}{99}\).Since differences are not constant, the sequence is not arithmetic.

Key Concepts

Sequence Formula

The sequence formula is a mathematical expression that tells you how to find the elements of a sequence based on the position. In our exercise, the given formula is \( a_n = \frac{1}{1+2n} \). This expression provides a rule that, when you plug in a position number \( n \), will give you the value of the sequence at that position.

• For example, if you want the first term of the sequence, replace \( n \) with 1, resulting in \( a_1 = \frac{1}{1+2(1)} = \frac{1}{3} \).
• Similarly, the second term can be found by setting \( n = 2 \), resulting in \( a_2 = \frac{1}{5} \).

This rule continues as you replace \( n \) with successive natural numbers to generate more terms in the sequence. Sequence formulas can differ greatly, depending on the type of sequence, such as arithmetic, geometric, or more complex ones.

Common Difference

The common difference is a crucial feature of an arithmetic sequence. This refers to the consistent number that you add to each term to get the next term. In previous attempts to determine the nature of a sequence, calculating this gives an indication of whether it's arithmetic.For a sequence to be arithmetic, the difference between successive terms must remain constant. In our example, we calculated the differences:

• \( a_2 - a_1 = \frac{1}{5} - \frac{1}{3} = -\frac{2}{15} \)
• \( a_3 - a_2 = \frac{1}{7} - \frac{1}{5} = -\frac{2}{35} \)
• \( a_4 - a_3 = \frac{1}{9} - \frac{1}{7} = -\frac{2}{63} \)
• \( a_5 - a_4 = \frac{1}{11} - \frac{1}{9} = -\frac{2}{99} \)

These differences are not the same, confirming that our sequence is not arithmetic. If the differences were equivalent, we would then say the sequence is arithmetic and recognize the common difference as the repeating value.

nth Term

The \( n \)th term of a sequence refers to the formula that describes the sequence's terms in general. It is usually expressed as \( a_n \). This allows you to find any specific term if you know its position \( n \) in the sequence.Consider the formula \( a_n = \frac{1}{1+2n} \), where \( n \) represents the position:

• Plug \( n = 1 \) to find the first term: \( a_1 = \frac{1}{3} \).
• For the second term, \( n = 2 \): \( a_2 = \frac{1}{5} \).
• This continues as you substitute different values of \( n \), like \( n = 3, 4, \) or \( 5 \), to get the corresponding terms.

By understanding the \( n \)th term, you're equipped to generate any term in the sequence, without calculating every preceding value. This is particularly beneficial in sequences with many terms, helping you leap to any part swiftly.