Problem 53 - Contemporary Precalculus

Question

Find the third and the sixth partial sums of the sequence. $$\left\\{(-1)^{n+1} 5\right\\}$$

Step-by-Step Solution

Answer

Answer: The third partial sum is 5 and the sixth partial sum is 0.

Step 1: Understand the sequence

The given sequence can be written as follows: $$a_n = (-1)^{n+1} 5$$ The sequence will alternate between positive and negative numbers due to the \((-1)^{n+1}\) factor. The terms of the sequence can also be written as: $$5, -5, 5, -5, 5, -5, \dots$$

Step 2: Calculate the third partial sum

To find the third partial sum, we need to calculate and sum the first three terms of the sequence: $$S_3 = a_1 + a_2 + a_3 = (-1)^2 5 + (-1)^3 5 + (-1)^4 5 = 5 - 5 + 5 = 5$$ So, the third partial sum is \(5\).

Step 3: Calculate the sixth partial sum

Similarly, to find the sixth partial sum, we need to calculate and sum the first six terms of the sequence: $$S_6 = a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = (-1)^2 5 + (-1)^3 5 + (-1)^4 5 + (-1)^5 5 + (-1)^6 5 + (-1)^7 5 = 5 - 5 + 5 -5 +5-5=0$$ So, the sixth partial sum is \(0\). In conclusion, the third partial sum of the sequence is \(5\) and the sixth partial sum is \(0\).

Key Concepts

Alternating Sequences

Alternating sequences are sequences in which the signs of the terms swap back and forth from positive to negative or vice versa. In mathematical terms, when dealing with such sequences, a common pattern is \((-1)^{n+1}\) or \((-1)^{n}\). This pattern helps to alternate the sign of each successive term. For a sequence formula like \((-1)^{n+1}5\), every term will flip from positive to negative or from negative to positive depending on whether \(n\) is even or odd.
This type of sequence is useful in various mathematical and real-world applications where periodic changes in direction or state are modeled.

• To recognize an alternating sequence, look for variables raised to a power of \(-1\) or examination of the sign pattern.
• Alternating sequences can impact the sums of terms, leading to potentially simpler calculations due to cancellations between positive and negative terms.

Learning to identify and work with alternating sequences can enhance understanding of complex series and sequences.

Arithmetic Calculations

Arithmetic calculations are fundamental to solving sequence-related problems. In the context of finding a partial sum in a sequence, these calculations involve performing simple addition and subtraction of terms. For the sequence \((-1)^{n+1} 5\), understanding how to calculate each term is crucial.

• Each term flip-flops its sign, affecting arithmetic calculations. For example, if \(n = 2\), \((-1)^{n+1}\) is positive, thus the term is \(5\).
• If \(n = 3\), \((-1)^{n+1}\) results in a negative value, making the term \(-5\).

When calculating partial sums, tally each term's value according to its position in the sequence.
To find the third partial sum, add up the first three terms, considering their signs: \(a_1 + a_2 + a_3 = 5 - 5 + 5\).
Similarly, the sixth partial sum requires adding six terms: \(5 - 5 + 5 - 5 + 5 - 5 = 0\). Mastering arithmetic calculations within sequences aids in gaining more confidence in handling larger and more complex series.

Sequence and Series Concepts

Sequences and series are essential components of mathematical studies. A sequence is a set of numbers arranged in a specific order, while a series is the sum of the terms of a sequence. In the context of partial sums, we aim to find the sum of a specific number of terms from the start of the sequence.

• A sequence can be finite or infinite, with infinite sequences continuing indefinitely.
• The concept of a partial sum involves summing up a certain number of initial terms in a sequence.
• For alternating sequences, the partial sum can reveal significant patterns due to the cancellation effect of positive and negative terms.

In exercises such as calculating the third and sixth partial sums of an alternating sequence, leveraging knowledge of sequences and series concepts streamlines the process. Understanding these underlying principles allows students to efficiently solve problems and accurately calculate results, fostering a deeper comprehension of mathematical patterns and operations.