Problem 42 - Algebra 2

Question

a. Open-Ended Write two explicit formulas for arithmetic sequences. b. Write the first five terms of each related series. c. Use summation notation to rewrite each series. d. Evaluate each series.

Step-by-Step Solution

Answer

The explicit formulas for the two arithmetic sequences are \(a_n = 2n - 1\) and \(a_n = 3n\). The first five terms of each series are 1, 3, 5, 7, 9 and 3, 6, 9, 12, 15 respectively. Summation notation for each series is \(\sum_{n=1}^{5} (2n - 1)\) and \(\sum_{n=1}^{5} 3n\) . The evaluations of each series are 25 and 45 respectively.

Step 1: Generating Arithmetic Sequences

Take two different common differences and starting points. For the first sequence, start at 1 with a common difference of 2: \(a_n = a_1 + (n-1)d = 1 + (n-1)2 = 2n - 1\). For the second sequence, start at 3 with a common difference of 3: \(a_n = a_1 + (n-1)d = 3 + (n-1)3 = 3n\).

Step 2: Creating the First Five Terms of the Series

Substitute the corresponding values of \(n = 1\), \(n = 2\), \(n = 3\), \(n = 4\), and \(n = 5\) into each of the sequences' formulas. For the first sequence, this results in 1, 3, 5, 7, 9. For the second sequence, the numbers 3, 6, 9, 12, 15 are obtained.

Step 3: Converting to Summation Notation

The first sequence sums from \(n = 1\) to \(n = 5\) for each term \(2n - 1\). This can be written in summation notation as \(\sum_{n=1}^{5} (2n - 1)\). The second sequence sums from \(n = 1\) to \(n = 5\) for each term \(3n\), which in summation notation appears as \(\sum_{n=1}^{5} 3n\).

Step 4: Evaluating the Series

The sum of the first sequence is calculated as: \(1 + 3 + 5 + 7 + 9 = 25\). The sum of the second sequence can be calculated as: \(3 + 6 + 9 + 12 + 15 = 45\).

Key Concepts

Explicit Formula

An explicit formula in arithmetic sequences allows us to find the nth term directly, without having to list all its preceding terms. It is of the form \(a_n = a_1 + (n-1)d\), where \(a_1\) is the first term and \(d\) is the common difference. This form is powerful as it gives a straightforward way to compute any term in the sequence. Let's see this in action: if we use a starting term \(a_1=1\) and a common difference \(d=2\), the explicit formula becomes \(a_n = 2n - 1\). This means that the nth term can be calculated directly by plugging any value of \(n\) into this formula.

Summation Notation

Summation notation provides a compact way to represent the addition of a sequence of numbers, using the sigma symbol \(\sum\). It essentially indicates that you are summing terms as defined by an equation for a particular range of terms. For instance, the sequence derived from the explicit formula \(2n - 1\) for \(n=1\) to \(5\) can be written using summation notation as \(\sum_{n=1}^{5} (2n - 1)\). This symbolizes the sum of terms from the first up to the fifth term, providing clarity and simplicity in written math.

Series Evaluation

Series evaluation is the process of adding up all the terms in a series. Once a sequence is written in summation notation, we then evaluate the entire sum by adding each individual term together. For example, with the summation \(\sum_{n=1}^{5} (2n - 1)\), you calculate each term for \(n = 1\) to \(5\), and then compute the total: \(1 + 3 + 5 + 7 + 9 = 25\). Through series evaluation, you find the total aggregation of sequence terms, often a notable element in mathematical analysis and problem-solving.

Common Difference

The common difference in arithmetic sequences is the value that separates each term from its successor. It is a crucial part of sequence identification and analysis. Every term is derived by adding this common difference to the previous term. For instance, in the sequence starting with \(1\) and having a common difference of \(2\), it forms \( 1, 3, 5, 7, 9\). Here, you see that \(2\) is added to each successive term, starting with \(1\). The common difference determines the uniformity and progression of an arithmetic sequence, allowing predictions and calculations to be made clear.