Problem 56 - Algebra 2

Question

Use a graphing calculator to find the sum of each geometric series. $$ \sum_{n=1}^{15} 2\left(\frac{1}{2}\right)^{n-1} $$

Step-by-Step Solution

Answer

The sum of the series is approximately 4.

Step 1: Identify the First Term and Common Ratio

In a geometric series, the first term \( a \) is the term when \( n = 1 \). Here, \( a = 2 \). The common ratio \( r \) is the factor by which each term is multiplied to get the next term. In this series, each subsequent term is found by multiplying the previous term by \( \frac{1}{2} \), so \( r = \frac{1}{2} \).

Step 2: Use the Formula for the Sum of a Geometric Series

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

Step 3: Substitute Known Values Into the Formula

In this problem, \( a = 2 \), \( r = \frac{1}{2} \), and \( n = 15 \). Substitute these values into the sum formula: \[ S_{15} = 2 \frac{1-(\frac{1}{2})^{15}}{1-\frac{1}{2}} \]

Step 4: Calculate \( r^n \)

Calculate \( (\frac{1}{2})^{15} \) using your graphing calculator, which equals approximately \( 3.0517578125 \times 10^{-5} \).

Step 5: Compute the Denominator

The denominator in the formula is \( 1 - \frac{1}{2} = \frac{1}{2} \).

Step 6: Compute the Numerator

The numerator is \( 1 - (\frac{1}{2})^{15} \), which calculates to approximately \( 1 - 3.0517578125 \times 10^{-5} = 0.999969482421875 \).

Step 7: Calculate the Sum \( S_{15} \)

Substitute the values into the formula: \[ S_{15} = 2 \times \frac{0.999969482421875}{0.5} \]. This results in approximately \( 3.9998779296875 \).

Step 8: Final Calculation

The sum of the series is approximately 4.

Key Concepts

Understanding the Common Ratio

The concept of a common ratio is fundamental in geometric series. It refers to the consistent factor by which each term in the series is multiplied to obtain the next term.
In the exercise at hand, the common ratio is denoted as \( r \) and has a value of \( \frac{1}{2} \). This means that to get from one term to the next, each term is halved.
This property creates a consistent pattern across the series, which is key for using formulas to solve problems such as finding the sum of terms. The identification of this common ratio is often the first step in analyzing any geometric series.

Applying the Sum Formula

To find the sum of the first \( n \) terms in a geometric series, we use a specific formula that requires us to know the first term \( a \), the common ratio \( r \), and the number of terms \( n \).
The formula is given by:

• \( S_n = a \frac{1-r^n}{1-r} \)

For our example, \( a = 2 \), \( r = \frac{1}{2} \), and \( n = 15 \).
By substituting these values into the formula, we determine how the terms in the series combine to give us a sum. This characterization simplifies complex series calculations and sheds light on the behavior of the entire sequence.

Using a Graphing Calculator

A graphing calculator can be a helpful tool in solving geometric series problems, especially when dealing with potentially cumbersome numerical computations.
It allows for quick calculation of exponentials, such as \( r^n \), which is crucial in our series sum formula.

• For instance, calculating \( \left(\frac{1}{2}\right)^{15} \) can be done instantly and accurately with a graphing calculator, giving a result of approximately \( 3.0517578125 \times 10^{-5} \).

These calculators can also handle other detailed steps, ensuring precision while sparing you from manual arithmetic that could lead to errors.

Exploring Exponential Calculations

Exponential calculations play a central role when dealing with geometric series. They arise from repeatedly applying the common ratio to the first term.
In our exercise, we computed \( \left(\frac{1}{2}\right)^{15} \).
This calculation determines how each term is scaled down exponentially as \( n \) increases, significantly impacting the term selection in the sum formula.
Exponential decay, for instance, means terms rapidly decrease in value, which makes their additions to the sum become negligible at some point.
Understanding these exponential behaviors helps predict the course of series progression, providing insights not just into mathematics, but into real-world applications where growth or decay follows similar patterns.