Problem 7 - Algebra 2

Question

Find the sum of each geometric series. $$ \sum_{n=1}^{7} 81\left(\frac{1}{3}\right)^{n-1} $$

Step-by-Step Solution

Answer

The sum of the series is 121.5.

Step 1: Identify the First Term

In a geometric series, the first term is denoted by \(a\). For the given series \(\sum_{n=1}^{7} 81\left(\frac{1}{3}\right)^{n-1}\), the first term \(a\) is 81 since at \(n=1, a\) is \(81 \left(\frac{1}{3}\right)^{0} = 81\).

Step 2: Identify the Common Ratio

The common ratio \(r\) in a geometric series is the factor by which we multiply each term to get the next term. From the expression, the common ratio \(r\) is \(\frac{1}{3}\).

Step 3: Determine the Number of Terms

The number of terms \(n\) in this series is given as 7, which is specified in the sum notation as \(\sum_{n=1}^{7}\).

Step 4: Use the Sum Formula for a Geometric Series

The sum of the first \(n\) terms of a geometric series can be calculated using the formula: \[ S_n = a \frac{1-r^n}{1-r} \]Substituting the known values, \(a = 81\), \(r = \frac{1}{3}\), and \(n = 7\), the formula becomes:\[ S_7 = 81 \frac{1 - \left(\frac{1}{3}\right)^7}{1-\frac{1}{3}} \]

Step 5: Calculate \(r^n\)

First, calculate \( \left(\frac{1}{3}\right)^7 \). This is \( \frac{1}{3^7} = \frac{1}{2187} \).

Step 6: Plug Values into Sum Formula

Substitute \( \left(\frac{1}{3}\right)^7 = \frac{1}{2187} \) into the sum formula:\[ S_7 = 81 \frac{1 - \frac{1}{2187}}{1 - \frac{1}{3}} \]

Step 7: Simplify the Denominator

The denominator of the fraction is \(1 - \frac{1}{3} = \frac{2}{3}\).

Step 8: Simplify the Numerator

Calculate the numerator, \(1 - \frac{1}{2187}\), to get \( \frac{2186}{2187} \).

Step 9: Compute the Sum

Substitute the simplified numerator and denominator back into the formula:\[ S_7 = 81 \times \frac{2186}{2187} \times \frac{3}{2} \]Continue simplifying:\[ S_7 = 81 \times \frac{6558}{4374} \]\[ S_7 = 81 \times 1.5 \approx 121.5 \]

Step 10: Conclusion

Thus, the sum of the geometric series is 121.5.

Key Concepts

First Term

In any geometric series, the first term is crucial as it sets the initial value from which the series begins. It's represented by the letter \( a \). Identifying the first term allows us to apply various formulas to find the sum or other properties of the series.
For instance, in the given geometric series \( \sum_{n=1}^{7} 81\left(\frac{1}{3}\right)^{n-1} \), the first term is easily determined by finding the term when \( n=1 \).
Here's how you can do it:

• Substitute \( n=1 \) into the expression: \( 81\left(\frac{1}{3}\right)^{1-1} \) becomes \( 81\left(\frac{1}{3}\right)^{0} \).
• Since \( \left(\frac{1}{3}\right)^{0} = 1 \), the first term \( a \) is \( 81 \times 1 = 81 \).

Knowing the first term helps in solidifying our understanding of the series structure and provides the base for further calculations.

Common Ratio

The common ratio is the factor by which each term in a geometric series is multiplied to get the next term. It's a constant ratio represented by \( r \). Understanding and identifying the common ratio is essential because it significantly influences the series' progression.
For example, in the series \( 81\left(\frac{1}{3}\right)^{n-1} \), the common ratio is \( \frac{1}{3} \). To spot it, notice the following pattern:

• The expression \( \left(\frac{1}{3}\right)^{n-1} \) explicitly shows that each term is multiplied by \( \frac{1}{3} \) as \( n \) increases by 1.

This constant multiplication means that each subsequent term is \( \frac{1}{3} \) of the previous term. Recognizing this ratio guides us in applying the sum formula and understanding the behavior of the series over its sequence.

Sum Formula for Geometric Series

Once you have identified both the first term and the common ratio, you can use the sum formula for a geometric series to find the sum of a specified number of terms.The sum formula is given by:\[ S_n = a \frac{1-r^n}{1-r} \]where \( S_n \) is the sum of the first \( n \) terms, \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.
To solve the given series \( \sum_{n=1}^{7} 81\left(\frac{1}{3}\right)^{n-1} \):

• Substitute the values \( a = 81 \), \( r = \frac{1}{3} \), and \( n = 7 \) into the formula.
• Calculate \( r^n = \left(\frac{1}{3}\right)^7 = \frac{1}{2187} \).
• Plug into the formula: \[ S_7 = 81 \frac{1 - \frac{1}{2187}}{1 - \frac{1}{3}} \]
• After simplifying, you find that \( S_7 = 121.5 \).

This formula is particularly powerful because it gives a quick way to calculate the sum without needing to manually add up each term. It shows the mathematical beauty and efficiency in understanding sequences.