Question

Find the sum of the series. $$ \sum_{k=0}^{6} 2\left(\frac{3}{4}\right)^{k} $$

Short answer

Question: Find the sum of the following geometric series: $$\sum_{k=0}^{6} 2\left(\frac{3}{4}\right)^{k}$$ Answer: The sum of the given geometric series is $$S_7 = \frac{113576}{16384}.$$

Step-by-step solution

Identify the given information

The geometric series is given as: $$ \sum_{k=0}^{6} 2\left(\frac{3}{4}\right)^{k} $$ We can identify that \(a = 2\left(\frac{3}{4}\right)^0 = 2\), the common ratio \(r = \frac{3}{4}\), and the total number of terms \(n = 7\) (since we start at \(k = 0\)).

Apply the geometric series formula

With the information given, we will apply the geometric series formula: $$S_n = a\left(\frac{1 - r^n}{1-r}\right)$$

Plug in the known values

Plugging in the values of \(a\), \(r\), and \(n\), we get: $$S_7 = 2\left(\frac{1 - (\frac{3}{4})^7}{1-\frac{3}{4}}\right)$$

Simplify the expression

After plugging in, simplify the expression: $$S_7 = 2\left(\frac{1 - (\frac{3}{4})^7}{\frac{1}{4}}\right)$$ $$S_7 = 2\left(4(1 - (\frac{3}{4})^7)\right)$$

Calculate the sum

Now, we need to calculate the value of the expression: $$S_7 = 2(4(1 - (2187/16384) ))$$ $$S_7 = 8(1 - (2187/16384))$$ $$S_7 = 8(\frac{16384 -2187}{16384})$$ $$S_7 = 8(\frac{14197}{16384})$$

Result

Therefore, the sum of the provided geometric series is: $$S_7 = \frac{113576}{16384}$$

Key concepts

Understanding the Common Ratio

In a geometric series, the common ratio is a key element. It defines how each term in the series relates to the one before it. Specifically, by multiplying the previous term by the common ratio, you obtain the next term in the series. This concept is fundamental for identifying and working with geometric progressions.

Let's delve into the exercise example. The series given is \( \sum_{k=0}^{6} 2\left(\frac{3}{4}\right)^{k} \). Here, the common ratio \( r \) is \( \frac{3}{4} \).

• Observe that multiplying any term by \( \frac{3}{4} \) will get you the next term.
• It's the constant factor that determines the progression rate through the series.

By understanding this, you can easily predict or calculate subsequent terms. Identifying the common ratio is the first step in leveraging the geometric series formula, which helps in finding particular series values, like their sums.

Sum of a Geometric Series

To find the sum of a geometric series involves adding up all its terms together. When each of these consecutive terms is defined by a constant multiple, it results in a specific pattern called a geometric progression. Calculating the total or sum of such series can be more straightforward when applying formulas.

For the exercise we looked at, the goal was to find out what sum we get when all the terms are added from the 0th to the 6th term. Therefore, it computes the sum of this series: \( \sum_{k=0}^{6} 2\left(\frac{3}{4}\right)^{k} \).

• It helps to start with identifying how many terms there are (here it is 7).
• Recognize the first term (denoted as \( a=2 \) at \( k = 0 \)). This simplifies the process.

Understanding how these terms grow and contribute to the total is crucial in applying the geometric series formula correctly.

The Geometric Series Formula

The geometric series formula is an essential tool for solving complex problems involving progression of numbers in a geometric series. It provides a standardized approach to calculating the sum efficiently.

The formula is given by \( S_n = a\left(\frac{1 - r^n}{1-r}\right) \), where:

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

Applying this formula simplifies calculating the sum over a large number of terms. In our exercise, by substituting \( a = 2 \), \( r = \frac{3}{4} \), and \( n = 7 \) into the formula, we calculated \( S_7 \), leading to a final sum of \( \frac{113576}{16384} \).

Grasping how the formula operates allows students to tackle similar series problems with confidence, utilizing it as a framework to reach accurate solutions efficiently.