Problem 45 - Contemporary Precalculus

Question

In Exercises \(43-48,\) find the sum. $$\sum_{n=1}^{9}\left(-\frac{1}{3}\right)^{n}$$

Step-by-Step Solution

Answer

Answer: The sum of the given geometric series is \(-\frac{229439}{1048576}\).

Step 1: Identify the first term, common ratio, and the number of terms

We can see that the series is a geometric series with 9 terms with a first term \(a_{1} \left(-\frac{1}{3}\right)^{1}\), and a common ratio \(r = -\frac{1}{3}\). The number of terms is \(n=9\).

Step 2: Apply the geometric series sum formula

The sum of a geometric series, given the first term (\(a_{1}\)), the common ratio (\(r\)) and the number of terms (\(n\)) can be found using the formula: $$S_{n} = \frac{a_{1}(1-r^{n})}{1-r}$$

Step 3: Calculate the sum

Now we can plug in the values into the formula and find the sum of the series: $$S_{9} = \frac{\left(-\frac{1}{3}\right)(1-(-\frac{1}{3})^{9})}{1-(-\frac{1}{3})}$$

Step 4: Simplify the expression

The expression can be simplified: $$S_{9} = \frac{-\frac{1}{3}(1-(-\frac{1}{3})^{9})}{\frac{4}{3}}$$ $$S_{9} = -\frac{1}{4}(1-(-\frac{1}{3})^{9})$$

Step 5: Evaluate the result

Now we can evaluate the result: $$S_{9} = -\frac{1}{4}(1-(-\frac{1}{3})^{9})$$ $$S_{9} = -\frac{1}{4}(1-\frac{262144}{19683})$$ $$S_{9} = -\frac{1}{4}(\frac{229439}{262144})$$ $$S_{9} = -\frac{229439}{1048576}$$ Therefore, the sum of the given series is: $$S_{9} = -\frac{229439}{1048576}$$

Key Concepts

Sum of Series

Understanding the concept of the sum of a series is essential when dealing with sequences, including geometric ones. A series is essentially the sum of the terms of a sequence. In our scenario, we're dealing with a geometric series. A geometric series is special because each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. To find the sum of this kind of series, we use a specific formula:

• Let \( a_1 \) be the first term of the series.
• Let \( r \) be the common ratio.
• \( n \) is the total number of terms.

The sum \( S_n \) of the first \( n \) terms of such a geometric series is given by the formula:\[S_n = \frac{a_1(1-r^n)}{1-r}\]This formula is particularly handy because it allows us to compute the sum without having to individually add up each term. This is especially useful when the series has many terms.
By substituting the correct values into this formula, we can find the sum of our given series quickly and accurately.

Geometric Sequence

A geometric sequence, which is sometimes referred to as a geometric progression, is a sequence where each term is derived from the previous one by multiplying it by a constant known as the common ratio.
For instance, consider a sequence where the first term is \( a \) and the common ratio is \( r \). The terms of the sequence are then given by:

• First term: \( a \)
• Second term: \( ar \)
• Third term: \( ar^2 \)
• and so on...

In the geometric sequence from our exercise, we start with \(-\frac{1}{3}\) as the first term and multiply by the common ratio \(-\frac{1}{3}\) to get each subsequent term. Each term is clearly dependent on its predecessor and the common ratio. Understanding this dependency is key to recognizing patterns in geometric sequences, which further simplifies the process of solving geometric series.

Common Ratio

The common ratio in a geometric sequence is the factor by which we multiply each term to get the next term. It plays a crucial role in defining how the sequence behaves. In a geometric sequence or series:

• If the common ratio \( r > 1 \), the terms will increase exponentially, growing larger and larger.
• If \( 0 < r < 1 \), the terms will decrease, getting smaller with each step.
• If \( -1 < r < 0 \), the terms will still decrease, but their signs will alternate.
• If \( r < -1 \), the terms will grow in magnitude, and the signs will alternate.

In our exercise, we were given a common ratio \( r = -\frac{1}{3} \). Because the value is between -1 and 0, the sequence experiences a diminishing magnitude with alternating signs.
The common ratio not only determines how each term in the sequence is generated but also heavily influences the sum of the entire series. By understanding the role of the common ratio, we can better comprehend and manipulate geometric sequences and series.