Problem 44 - Precalculus Mathematics for Calculus

Question

Find the partial sum \(S_{n}\) of the geometric sequence that satisfies the given conditions. $$a=\frac{2}{3}, \quad r=\frac{1}{3}, \quad n=4$$

Step-by-Step Solution

Answer

The partial sum \( S_4 \) is \( \frac{80}{81} \).

Step 1: 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.

Step 2: 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: \( S_4 = \frac{2}{3} \frac{1-(\frac{1}{3})^4}{1-\frac{1}{3}} \).

Step 3: Simplify the Exponent

Calculate \((\frac{1}{3})^4\), which equals \(\frac{1}{81}\). So, the expression updates to: \( S_4 = \frac{2}{3} \frac{1-\frac{1}{81}}{1-\frac{1}{3}} \).

Step 4: Calculate the Denominator

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

Step 5: Calculate the Numerator

Subtract \(\frac{1}{81}\) from 1, giving \(1 - \frac{1}{81} = \frac{80}{81}\).

Step 6: Solve the Fraction

Perform the division in the formula: \(\frac{80}{81} \div \frac{2}{3} = \frac{80}{81} \times \frac{3}{2} = \frac{240}{162}\).

Step 7: Simplify the Fraction

Simplify \(\frac{240}{162}\) by finding the greatest common divisor, which is 6. Thus, \(\frac{240}{162} = \frac{40}{27}\).

Step 8: Final Calculation of Partial Sum

Finally, multiply the simplified fraction \(\frac{40}{27}\) by the first term \(\frac{2}{3}\): \( S_4 = \frac{2}{3} \times \frac{40}{27} = \frac{80}{81}\).

Key Concepts

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:

• \(a,\) 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.

Using these values, the formula calculates how much "in total" a specified section of the sequence adds up to by accounting for how each term exponentially builds on the last. For students, mastering this formula involves understanding each variable's role and how they interact together in the equation.

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:

• **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.

By understanding the common ratio, students can predict how the sequence behaves over time and consequently how the partial sum evolves.

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:

• **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.

A firm grasp of exponential calculations is vital to effectively utilize the partial sum formula as it involves raising the common ratio to the power of \( n \), integral for accurate results.