Problem 14 - College Algebra with Modeling and Visualization

Question

For the given \(a_{n},\) calculate \(S_{5}.\) $$ a_{n}=\frac{1}{2 n} $$

Step-by-Step Solution

Answer

The sum \(S_5 = \frac{137}{120}.\)

Step 1: Understand the Problem

We need to find the sum of the first 5 terms of the sequence \(a_n\) where \(a_n = \frac{1}{2n}\). This is denoted as \(S_5 = a_1 + a_2 + a_3 + a_4 + a_5\).

Step 2: Calculate Individual Terms

Find the first 5 terms of the sequence by substituting \(n = 1\) to \(n = 5\) into the formula \(a_n = \frac{1}{2n}\). \[a_1 = \frac{1}{2\times1} = \frac{1}{2}\]\[a_2 = \frac{1}{2\times2} = \frac{1}{4}\]\[a_3 = \frac{1}{2\times3} = \frac{1}{6}\]\[a_4 = \frac{1}{2\times4} = \frac{1}{8}\]\[a_5 = \frac{1}{2\times5} = \frac{1}{10}\]

Step 3: Add the Terms

Add the terms calculated in the previous step to find \(S_5\).\[S_5 = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10}\]

Step 4: Simplify the Sum

Convert each fraction to a common denominator and add them together: \[S_5 = \frac{60}{120} + \frac{30}{120} + \frac{20}{120} + \frac{15}{120} + \frac{12}{120} = \frac{137}{120}\] Therefore, \(S_5 = \frac{137}{120}.\)

Key Concepts

Arithmetic series

An arithmetic series is a sequence of numbers in which each term after the first is formed by adding a fixed, repeated value, called the common difference, to the previous term. Understanding arithmetic series is essential as they frequently appear in various mathematical problems and real-world applications. Unlike a general sequence, an arithmetic series has this uniformity in difference, making calculations and analysis straightforward.

While the exercise we are analyzing is not an arithmetic series, understanding arithmetic series helps differentiate them from other types of sequences like geometric series or harmonic series. For example, if you have a series like 2, 4, 6, 8,..., where each term increases by 2, then this sequence is categorized as an arithmetic series with a common difference of 2.

• Arithmetic series are linear and predictable.
• These simplify the process of finding the sum by using a formula rather than adding each element sequentially.
• The formula for an arithmetic series is useful: if the first term is \(a\), and the last term is \(l\), the sum \(S_n\) of the first \(n\) terms can be found by \(S_n = \frac{n}{2} (a + l)\).

Sum of sequence

The sum of a sequence refers to the total value obtained by adding all terms of the sequence together. Finding this sum can sometimes be tricky, especially if the sequence is long or the terms are complex.

In the problem, calculating the sum of the first 5 terms involves the sequence \(a_n = \frac{1}{2n}\). Here, each term is a fraction. To discover the sum, each fraction was added together: \[S_5 = \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10}\] When dealing with arithmetic sequences, finding the sum is often simplified with formulas, as described before. Yet for this unique sequence, one must calculate each term and then perform the addition manually. However, an important step involves handling fractions properly, which leads to the next significant concept.

Fraction simplification

Fraction simplification is a crucial mathematical process where fractions are reduced to their simplest form. This concept frequently appears when dealing with sequences and series, as it aids in making sums more manageable and results more understandable.

In solving this exercise, the sum of fractions required finding a common denominator. By simplifying, we transformed:

• \(\frac{1}{2}\) into \(\frac{60}{120}\)
• \(\frac{1}{4}\) into \(\frac{30}{120}\)
• \(\frac{1}{6}\) into \(\frac{20}{120}\)
• \(\frac{1}{8}\) into \(\frac{15}{120}\)
• \(\frac{1}{10}\) into \(\frac{12}{120}\)

Then add these simplified fractions:\[S_5 = \frac{60}{120} + \frac{30}{120} + \frac{20}{120} + \frac{15}{120} + \frac{12}{120} = \frac{137}{120}\]Fraction simplification helps to standardize various fractions, allowing easy addition. This results in reduced calculation steps and avoiding errors that might occur with improper fractions. Mastery of fraction simplification greatly eases computational burdens across mathematical tasks.