Chapter 12: Problem 44
Find the partial sum \(S_{n}\) of the geometric sequence that satisfies the given conditions. $$a=\frac{2}{3}, \quad r=\frac{1}{3}, \quad n=4$$
Short Answer
Expert verified
The partial sum \( S_4 \) is \( \frac{80}{81} \).
Step by step solution
01
Identify the Formula for Partial Sum
The formula for the partial sum of a geometric sequence 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.
02
Substitute Known Values
Substitute the given values into the formula: \(a = \frac{2}{3}\), \(r = \frac{1}{3}\), and \(n = 4\). This leads to the expression: \( S_4 = \frac{2}{3} \frac{1-(\frac{1}{3})^4}{1-\frac{1}{3}} \).
03
Simplify the Exponent
Calculate \((\frac{1}{3})^4\), which equals \(\frac{1}{81}\). So, the expression updates to: \( S_4 = \frac{2}{3} \frac{1-\frac{1}{81}}{1-\frac{1}{3}} \).
04
Calculate the Denominator
The denominator of the fraction within the formula becomes \(1 - \frac{1}{3} = \frac{2}{3}\).
05
Calculate the Numerator
Subtract \(\frac{1}{81}\) from 1, giving \(1 - \frac{1}{81} = \frac{80}{81}\).
06
Solve the Fraction
Perform the division in the formula: \(\frac{80}{81} \div \frac{2}{3} = \frac{80}{81} \times \frac{3}{2} = \frac{240}{162}\).
07
Simplify the Fraction
Simplify \(\frac{240}{162}\) by finding the greatest common divisor, which is 6. Thus, \(\frac{240}{162} = \frac{40}{27}\).
08
Final Calculation of Partial Sum
Finally, multiply the simplified fraction \(\frac{40}{27}\) by the first term \(\frac{2}{3}\): \( S_4 = \frac{2}{3} \times \frac{40}{27} = \frac{80}{81}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Sum
In a geometric sequence, the partial sum refers to the sum of a finite number of consecutively listed terms. Understanding the concept of partial sum is essential for solving various problems involving geometric sequences. A geometric sequence itself is identified by its constant ratio between consecutive terms.
The formula to calculate the partial sum, denoted as \( S_n \), is \( S_n = a \frac{1-r^n}{1-r} \). Here:
The formula to calculate the partial sum, denoted as \( S_n \), is \( S_n = a \frac{1-r^n}{1-r} \). Here:
- \(a,\) the first term, sets the starting point of the sequence.
- \(r,\) the common ratio, is the factor by which we multiply one term to get to the next.
- \(n,\) the number of terms, indicates how many terms we want to include in our sum.
Common Ratio
The common ratio is a crucial characteristic of any geometric sequence. It is expressed as \( r \) and helps determine how one term in the sequence is related to the next. In our example, the common ratio was given as \( \frac{1}{3} \). This means every term is obtained by multiplying the previous term by \( \frac{1}{3} \).
The importance of the common ratio can be summarized as follows:
The importance of the common ratio can be summarized as follows:
- **Uniformity:** It keeps the sequence consistent, as each term is generated from the previous one by a constant multiplying factor.
- **Behavior prediction:** A ratio greater than 1 indicates the series will grow, while a ratio less than 1 (but greater than 0) suggests the terms decrease.
- **Sign direction:** If the ratio is negative, the signs of terms will alternate.
Exponential Calculation
Exponential calculation is a key process in determining terms within a geometric sequence. Since each term of the sequence can be expressed as \( ar^{n-1} \), the role of exponential calculation is apparent as we compute \( r^n \) for terms and for the partial sum formula.
Let's dissect the use of exponential calculations in our example:
Let's dissect the use of exponential calculations in our example:
- **Calculation of powers:** Here, we calculated \( \left(\frac{1}{3}\right)^4 = \frac{1}{81} \). Understanding powers of fractions helps in effectively evaluating the growth or decay pattern of sequences.
- **Simplifying expressions:** Exponents can simplify otherwise complex multiplications, breaking down large calculations into manageable components.