Problem 45 - Precalculus: Mathematical for Calculus

Question

Find the sum. $$\sum_{k=0}^{10} 3\left(\frac{1}{2}\right)^{k}$$

Step-by-Step Solution

Answer

The sum of the series is \( \frac{12282}{2048} \).

Step 1: Recognize the Series Type

The series given is a geometric series. A geometric series is of the form \( a \, r^n \), where \( a \) is the first term, and \( r \) is the common ratio.

Step 2: Identify the First Term and Common Ratio

In this series, the first term \( a \) is \( 3 \cdot \left(\frac{1}{2}\right)^0 = 3 \), and the common ratio \( r \) is \( \frac{1}{2} \).

Step 3: Use the Formula for the Sum of a Finite Geometric Series

The sum \( S_n \) of the first \( n+1 \) terms of a geometric series is given by the formula \[ S_n = a \frac{1-r^{n+1}}{1-r} \]where \( n \) is the number of terms minus one. Here, \( n = 10 \).

Step 4: Substitute the Values into the Formula

Using the formula from Step 3, substitute \( a = 3 \), \( r = \frac{1}{2} \), and \( n = 10 \):\[ S_{10} = 3 \frac{1 - \left(\frac{1}{2}\right)^{11}}{1 - \frac{1}{2}} \]

Step 5: Simplify the Expression

Calculate \( \left(\frac{1}{2}\right)^{11} \) and then substitute all values into the formula.\[ \left(\frac{1}{2}\right)^{11} = \frac{1}{2048} \]\[ S_{10} = 3 \frac{1 - \frac{1}{2048}}{\frac{1}{2}} \]

Step 6: Final Calculation

Simplify the expression to find the sum:\[ S_{10} = 3 \times 2 \left(1 - \frac{1}{2048}\right) \]\[ S_{10} = 6 \left(1 - \frac{1}{2048}\right) \]\[ S_{10} = 6 \times \frac{2047}{2048} \]\[ S_{10} = \frac{6 \times 2047}{2048} = \frac{12282}{2048} \]Since the number cannot be simplified further, this is the final sum of the series.

Key Concepts

Sum of a Geometric Series

A geometric series is a fascinating type of sequence where each term is obtained by multiplying the previous term by a constant value known as the common ratio. The general form of this series is represented as \(a, ar, ar^2, ar^3, \ldots\). The sum of a geometric series is all about adding up these terms.

To better understand, picture the series \(3, \frac{3}{2}, \frac{3}{4}, \ldots\) where each term is half of the one before. The first term in our example is 3, and we repeatedly multiply by \(\frac{1}{2}\) to get successive terms.

• The series given starts at the 0th power, making it a slightly simpler resolution.
• For a finite geometric series, we can sum up all the terms to a certain point.

Finite Geometric Series Formula

To find the sum of a finite geometric series, we rely on a crucial formula. This formula makes the calculation much easier no matter how long the series goes.
The formula for the sum of a finite geometric series is given by: \[ S_n = a \frac{1 - r^{n+1}}{1 - r} \] Where:

• \(a\) is the first term of the series.
• \(r\) is the common ratio, which in our case is \(\frac{1}{2}\).
• \(n\) is the number of terms minus one, so for 11 terms you use \(n=10\).

This formula is derived from the way each term in a geometric sequence relates multiplicatively. By harnessing these relationships, we avoid the tedious process of adding each term individually, especially when there are many terms.

Using the formula, we substitute the known values, and the calculation becomes a few straightforward steps.

Series Calculation

Once we have identified the correct series and know the formula, the calculation is all about plugging in those numbers correctly.
In the context of our sum, substituting the values \(a = 3\), \(r = \frac{1}{2}\), and \(n = 10\) into the formula gives us:\[ S_{10} = 3 \frac{1 - \left(\frac{1}{2}\right)^{11}}{1 - \frac{1}{2}} \] Now let's break this calculation down step by step:

• Calculate \(\left(\frac{1}{2}\right)^{11} = \frac{1}{2048}\).
• Substitute back the calculated power into the formula: \[ S_{10} = 3 \frac{1 - \frac{1}{2048}}{\frac{1}{2}} \]
• Continue by simplifying this to find the outcome: \[ S_{10} = 6 \times \frac{2047}{2048} \]
• Perform final multiplication to obtain the sum: \[ \frac{12282}{2048} \]

This brings us to a final result for the series, as the steps unfold naturally into a clear calculation. Remember, practice with these steps infuses confidence in tackling other series problems!