Problem 58 - Algebra 2

Question

Use a graphing calculator to find the sum of each geometric series. $$ \sum_{n=1}^{13} 6\left(\frac{1}{3}\right)^{n-1} $$

Step-by-Step Solution

Answer

The sum of the series is approximately 9.

Step 1: Understanding the Problem

We are given a geometric series \( \sum_{n=1}^{13} 6\left(\frac{1}{3}\right)^{n-1} \). We need to find its sum. The series has the first term 6 and a common ratio of \( \frac{1}{3} \).

Step 2: Identify Formula for Geometric Series

We use the formula for the sum of the first \( n \) terms of a geometric series: \( S_n = a\frac{1-r^n}{1-r} \), where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

Step 3: Substitute the Values into the Formula

Here, \( a = 6 \), \( r = \frac{1}{3} \), and \( n = 13 \). Substituting these values, we have \( S_{13} = 6\frac{1-\left(\frac{1}{3}\right)^{13}}{1-\frac{1}{3}} \).

Step 4: Calculate the Series Sum

Simplify and calculate: \( S_{13} = 6 \times \frac{1-\left(\frac{1}{3}\right)^{13}}{\frac{2}{3}} \). Simplify further to \( S_{13} = 6 \times \frac{3}{2} \times \left(1 - \left(\frac{1}{3}\right)^{13}\right) \).

Step 5: Use Graphing Calculator for Final Calculation

Use a graphing calculator to find \( \left(\frac{1}{3}\right)^{13} \), then substitute back to find: \( S_{13} = 6 \times \frac{3}{2} \times \left(1 - x\right) \) where \( x \) is the value of \( \left(\frac{1}{3}\right)^{13} \). Calculate the result to get \( S_{13} \approx 9 \).

Key Concepts

graphing calculator

A graphing calculator is a powerful tool that can help you visualize and solve complex math problems, such as geometric series. It allows you to input the series into the calculator and see the graphically represented terms. This makes it easier to understand how the terms trend and accumulate.

For a geometric series, you can input the function representing the series into the calculator to find specific values or plot the terms. In our specific case, the series is:

• \( 6\left(\frac{1}{3}\right)^{n-1} \) from \( n = 1 \) to \( n = 13 \).

By using the calculator, you can calculate complex expressions, such as \( \left(\frac{1}{3}\right)^{n} \) for large values like \( n = 13 \), which might be cumbersome to do manually. Additionally, for series calculations, you can directly use it to sum terms, thus obtaining quick results for problems requiring multiple computations.

sum of series formula

The sum of a geometric series is found using the sum of series formula. This formula helps you determine the cumulative value of the series over a specific number of terms. It's written as:

• \( S_n = a\frac{1-r^n}{1-r} \)

Where:

• \( S_n \) is the sum of the series.
• \( a \) is the first term of the series.
• \( r \) is the common ratio.
• \( n \) is the number of terms.

By substituting the series values into this formula, we calculate the sum efficiently.

In our example:

• \( a = 6 \)
• \( r = \frac{1}{3} \)
• \( n = 13 \)

Using the formula, we substitute these values to find the total sum for 13 terms.

common ratio

In a geometric series, the common ratio is a constant factor that each term is multiplied by to arrive at the next term. Understanding this ratio is key to analyzing the series' behavior. For our series, the common ratio is \( \frac{1}{3} \).

This value tells us that each subsequent term in the series is one-third of the previous term. The common ratio can dramatically affect the growth or decay of the series. If \( |r| < 1 \), such as in this case with \( r = \frac{1}{3} \), the series terms will decrease in size, leading the series towards a limit, giving the series a converging nature.

• If \( r \) is less than one, the terms get smaller.
• If \( r \) is greater than one, the terms get larger.

This insight is useful when using the sum of series formula or setting up calculations on a graphing calculator.

first term

The first term in a geometric series is important as it sets the starting point from which all other terms are derived using the common ratio. For the given geometric series, the first term \( a \) is 6.

This value initiates the sequence and plays a critical role in the sum of series formula. Each subsequent term is obtained by multiplying the previous term by the common ratio \( \frac{1}{3} \). Therefore, the entire pattern of the series is based on this first term and the common ratio.

• Knowing the first term helps determine the base value before ratio adjustments.
• Each calculation for the series setup requires this starting value.

Hence, understanding the first term is essential for calculations involving any geometric series.