Problem 60 - Precalculus: Mathematics for Calculus

Question

Find the sum. $$\sum_{k=1}^{5} 8\left(-\frac{3}{2}\right)^{k-1}$$

Step-by-Step Solution

Answer

The sum is 27.5.

Step 1: Identify the Type of Series

The given expression is a geometric series represented by the formula \( a r^{k-1} \) where \( a = 8 \) and \( r = -\frac{3}{2} \). This series starts from \( k = 1 \) and ends at \( k = 5 \).

Step 2: Write the Geometric Series Formula

The sum of the first \( n \) terms of a geometric series is given by the formula:\[ S_n = a \frac{1 - r^n}{1 - r} \]where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms.

Step 3: Substitute the Values into the Formula

Here, \( a = 8 \), \( r = -\frac{3}{2} \), and \( n = 5 \). Substitute these into the sum formula:\[ S_5 = 8 \frac{1 - \left(-\frac{3}{2}\right)^5}{1 + \frac{3}{2}} \]

Step 4: Compute \((-\frac{3}{2})^5\)

Calculate \((-\frac{3}{2})^5\):\[ (-\frac{3}{2})^5 = -\left(\frac{3^5}{2^5}\right) = -\frac{243}{32} \].

Step 5: Calculate the Denominator

Calculate \(1 + \frac{3}{2}\):\[ 1 + \frac{3}{2} = \frac{2}{2} + \frac{3}{2} = \frac{5}{2} \].

Step 6: Substitute and Simplify

Substitute \((-\frac{3}{2})^5 = -\frac{243}{32}\) and denominator value into the formula:\[ S_5 = 8 \frac{1 - ( -\frac{243}{32} )}{\frac{5}{2}} \].\[ S_5 = 8 \frac{\frac{32}{32} + \frac{243}{32}}{\frac{5}{2}} \].

Step 7: Simplify the Numerator

Calculate the numerator:\[ \frac{32 + 243}{32} = \frac{275}{32} \].

Step 8: Simplify the Entire Expression

Insert the new numerator and resolve the division by \( \frac{5}{2} \):\[ S_5 = 8 \frac{\frac{275}{32}}{\frac{5}{2}} \]Multiply by the reciprocal:\[ S_5 = 8 \times \frac{275}{32} \times \frac{2}{5} \].

Step 9: Perform Final Calculations

Carry out the multiplication:\[ S_5 = 8 \times \frac{275}{32} \times \frac{2}{5} = \frac{4400}{160} = 27.5 \].

Key Concepts

Sum of Series

In mathematics, finding the sum of a series involves adding up all the terms in the sequence. A geometric series is a type of series that significantly simplifies this process due to its consistent pattern.
Each term in a geometric series is obtained by multiplying the previous term by a constant value, known as the common ratio. Therefore, if we need to find the sum of a specified number of terms, we can apply a specific formula rather than adding each term separately.
This simplification saves time and minimizes the risk of errors in calculations, making it very convenient when dealing with long sequences.

Common Ratio

The common ratio in a geometric series is a crucial element. It is represented as the factor by which each term is multiplied to get the next term.
For example, if the terms of a series are generated by multiplying each preceding term by \(-\frac{3}{2}\), then \(-\frac{3}{2}\) is the common ratio.
This consistent multiplier gives geometric series their unique properties and enables us to predict subsequent terms when the first term and common ratio are known.
In the given exercise, understanding this concept allows us to calculate the sum efficiently by applying the sum formula.

Geometric Progression

Geometric progression, also known as a geometric sequence, is where the ratio between successive terms remains constant.
This sequence is one of the core topics of precalculus and is fundamental in solving real-world problems where exponential growth or decay is involved.
Each term in the sequence is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio.
In the exercise, the progression from term to term followed the pattern established by the formula \(a r^{k-1}\).
Being able to recognize and apply geometric progression is essential for students as it lays the groundwork for more complex mathematical concepts.

Precalculus

Precalculus serves as the foundational groundwork for understanding concepts in calculus. It encompasses a range of topics, including functions, series, and sequences, one of which is the geometric series.
This branch of mathematics prepares students for the abstract and dynamic nature of calculus, making sure they have a solid understanding of limits, continuity, and the behavior of functions.
By studying topics like geometric progression and the sum of a series, students develop problem-solving skills that are crucial, especially when they progress to calculus and beyond.
Grasping these early concepts ensures that students can tackle more advanced mathematics with confidence and precision.