Problem 33 - A Graphical Approach to Precalculus with Limits

Question

Use a formula to find the sum of each series. $$\sum_{j=1}^{6} 48\left(\frac{1}{2}\right)^{j}$$

Step-by-Step Solution

Answer

The sum of the series is 94.5.

Step 1: Identify the Series Type

The series given is a geometric series because the terms are created by multiplying a constant factor of \( \frac{1}{2} \) each time. We can identify that the first term \( a \) is 48, and the common ratio \( r \) is \( \frac{1}{2} \).

Step 2: Use the Geometric Series Sum Formula

The formula for the sum of the first \( n \) terms of a geometric series is given by:\[ S_n = a \frac{1 - r^n}{1 - r} \]In this problem, \( n = 6 \), \( a = 48 \), and \( r = \frac{1}{2} \).

Step 3: Substitute Known Values into the Formula

Substitute the known values into the geometric series sum formula:\[ S_6 = 48 \frac{1 - \left(\frac{1}{2}\right)^6}{1 - \frac{1}{2}} \]

Step 4: Simplify the Expression

First, calculate \( \left(\frac{1}{2}\right)^6 \) which gives \( \frac{1}{64} \). Now substitute this value into the equation:\[ S_6 = 48 \frac{1 - \frac{1}{64}}{1 - \frac{1}{2}} = 48 \frac{63/64}{1/2} \]

Step 5: Calculate the Sum

Since dividing by \( \frac{1}{2} \) is the same as multiplying by 2, simplify as follows:\[ S_6 = 48 \times 2 \times \frac{63}{64} = 48 \times \frac{126}{64} \]Calculating this gives:\[ S_6 = \frac{6048}{64} = 94.5 \]

Step 6: State the Final Answer

The sum of the series \( \sum_{j=1}^{6} 48\left(\frac{1}{2}\right)^{j} \) is 94.5.

Key Concepts

Series Sum Formula

The series sum formula is a vital tool when dealing with geometric series. It allows us to find the sum of a sequence of numbers that form a pattern. In the context of geometric series, where each term is derived by multiplying the previous term by a fixed value, the formula simplifies the process significantly.

For a geometric series, the sum of the first \( n \) terms is calculated using the formula:

• \( S_n = a \frac{1 - r^n}{1 - r} \)

where:

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

To use this formula efficiently, simply determine these values for the series you are working with and substitute them into the formula.

Common Ratio

The common ratio is a crucial element in understanding geometric sequences and series. It is the constant factor that each term is multiplied by to get the next term in the sequence. Identifying the common ratio helps in recognizing the kind of series you are dealing with as well as calculating terms and sums.

In the given exercise, the common ratio is \( \frac{1}{2} \). This means each subsequent term is half the previous term. Identifying this ratio enabled us to apply the geometric series sum formula correctly. It simplifies determining not only the following terms but also the cumulative sum of several terms in the series.

Geometric Sequence

A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed number called the common ratio. This concept is fundamental in understanding series as it provides the structural basis of calculations.

For the exercise, the sequence consists of terms with a starting value of 48. Each subsequent term is 48 multiplied by \( \left(\frac{1}{2}\right)^j \), where \( j \) ranges from 1 to 6. This dependency on a common ratio defines the sequence as geometric. Understanding this allows you to anticipate future values or use the series sum formula to find the total of the terms.

Precalculus

Precalculus serves as the bridge between algebra and calculus. It involves the study of concepts and tools essential for the calculus you'll encounter later in more advanced mathematics.

When dealing with topics such as geometric sequences and series, precalculus provides the necessary foundation. It equips students with the knowledge to use formulas like the series sum formula and offers insight into the behavior of sequences. Mastering these precalculus concepts is crucial, as they underpin a substantial portion of calculus, particularly in studying limits, derivatives, and integrals.