Problem 40 - A Graphical Approach to Precalculus with Limits

Question

Use a formula to find the sum of each series. $$\sum_{j=2}^{7} \frac{1}{3}(4)^{j-1}$$

Step-by-Step Solution

Answer

The sum of the series is 1820.

Step 1: Understand the Formula for a Geometric Series

The formula for the sum of a finite geometric series \( S_n \) 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 2: Identify the Series Components

First, let's identify the components of our series:- The series is defined by \( \sum_{j=2}^{7} \frac{1}{3}(4)^{j-1} \).- The first term \( a \) is found by substituting \( j=2 \) into the formula: \( a = \frac{1}{3}(4)^{2-1} = \frac{1}{3}(4)^1 = \frac{4}{3} \).- The common ratio \( r \) is 4, as seen in the expression \( 4^{j-1} \).- For the sum from \( j=2 \) to \( j=7 \), there are \( n = 7 - 2 + 1 = 6 \) terms.

Step 3: Calculate the Sum

Substitute the values into the formula for the sum of the geometric series:\[ S_6 = \frac{4}{3} \frac{1-4^6}{1-4} \]Calculate each part:- Compute \( 4^6 = 4096 \).- Then calculate \( 1-4^6 = 1-4096 = -4095 \).- Substitute these into the formula:\[ S_6 = \frac{4}{3} \frac{-4095}{-3} = \frac{4}{3} \times 1365 = 1820 \]

Step 4: Conclusion

The sum of the series \( \sum_{j=2}^{7} \frac{1}{3}(4)^{j-1} \) is 1820.

Key Concepts

finite series

A finite series is a summation of a finite number of terms. Unlike an infinite series that continues indefinitely, a finite series has a specific number of terms. This series allows us to calculate the sum with certainty. In mathematical problems, especially when dealing with geometric series like the one in our example, it is crucial to recognize whether it is finite.

• Clear boundaries: For a finite series, we know exactly where it begins and ends.
• Efficiency: Knowing the exact number of terms makes it easier to handle and compute the series.

For example, in the series \( \sum_{j=2}^{7} \frac{1}{3}(4)^{j-1} \), the calculation is only from \( j = 2 \) to \( j = 7 \). Hence, we calculate just six terms.
Understanding the limits of the series enables us to apply the formula correctly and avoid any errors that might arise due to an assumption of infinite terms.

common ratio

The common ratio in a geometric sequence or series is a fixed, constant factor between consecutive terms.
In simple words, it defines how each term is obtained from the previous one by multiplying by this fixed number. If the series is written in the form of a sequence, the common ratio is consistent throughout.
In our exercise, the common ratio (\( r \)) is found in the expression \( 4^{j-1} \). It is \( 4 \), meaning each term is four times the previous term.

• Identification: Spotting the common ratio swiftly lets you apply the geometric formula seamlessly.
• Consistency: The common ratio helps maintain the integrity and pattern of the sequence.

By understanding the common ratio, computation becomes effortless as it provides a clear multiplication pattern, forming the heart of geometric changes in the sequence.

geometric sequences

A geometric sequence is a series of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
This pattern of multiplication is what differentiates geometric sequences from other types of sequences.
In examining the series \( \sum_{j=2}^{7} \frac{1}{3}(4)^{j-1} \), each term is derived by this consistent multiplication:

• First term \( a \): Calculated using \( j = 2 \) here is \( \frac{4}{3} \)
• Common pattern: Moving from term to term involves multiplying by \( 4 \).

Geometric sequences are not only beautiful in their symmetry but are also incredibly practical. They often model exponential growth or decay in real-world scenarios, such as compound interest, populations, and radioactive decay. Understanding the concept of geometric sequences equips students with tools to approach a wide variety of mathematical problems.