Problem 47 - Beginning and Intermediate Algebra

Question

Evaluate each series. $$\sum_{i=1}^{5}(-1)^{i+1} \cdot(i)$$

Step-by-Step Solution

Answer

The final result of the series \(\sum_{i=1}^{5}(-1)^{i+1} \cdot(i) = 1 - 2 + 3 - 4 + 5 = 4\).

Step 1: Identify the range of the summation

The given series has a range from i=1 to i=5. We will substitute each of these values in the formula.

Step 2: Substitute each value of i in the formula

We will now substitute each value of i from 1 to 5 in the formula and calculate the terms as follows: - When \(i=1\), the term is: \((-1)^{1+1} \cdot(1) = (-1)^2 \cdot 1 = 1\) - When \(i=2\), the term is: \((-1)^{2+1} \cdot(2) = (-1)^3 \cdot 2 = -2\) - When \(i=3\), the term is: \((-1)^{3+1} \cdot(3) = (-1)^4 \cdot 3 = 3\) - When \(i=4\), the term is: \((-1)^{4+1} \cdot(4) = (-1)^5 \cdot 4 = -4\) - When \(i=5\), the term is: \((-1)^{5+1} \cdot(5) = (-1)^6 \cdot 5 = 5\)

Step 3: Sum up the terms

Now that we have found the value of each term, we will sum them up to get the result of the series: $$\sum_{i=1}^{5}(-1)^{i+1} \cdot(i) = 1 - 2 + 3 - 4 + 5 $$

Step 4: Calculate the sum

Finally, we will calculate the sum of the terms: $$1 - 2 + 3 - 4 + 5 = (-1) + 1 - 1 + 5 = 4$$

Step 5: Answer

So the final result of the series \(\sum_{i=1}^{5}(-1)^{i+1} \cdot(i)\) is 4.

Key Concepts

Arithmetic Series

An Arithmetic series is a sequence of numbers in which the difference between consecutive terms is constant. This difference is known as the common difference. For example, in the series \(2, 4, 6, 8\), the common difference is 2. In general, an arithmetic series can be written as the sum of the terms: \(a, a+d, a+2d, \ldots, a+(n-1)d\), where \(a\) is the first term, \(d\) is the common difference, and \(n\) is the number of terms.

To find the sum \(S_n\) of the first \(n\) terms of an arithmetic series, you can use the formula:\[ S_n = \frac{n}{2} [2a + (n - 1) d] \] Here, \(n\) is the number of terms to add together, \(a\) is the first term, and \(d\) is the common difference.

Arithmetic series are straightforward to understand because you can figure out the sum by simply understanding the pattern of adding a constant number repeatedly.

Alternating Series

An alternating series is a series where the signs of the terms alternate between positive and negative. This results in a sequence that oscillates about zero or another finite value. A simple example includes the series \(1, -2, 3, -4, 5, \ldots\) where each alternate term changes sign.

In the exercise, the series \( \sum_{i=1}^{5} (-1)^{i+1} \cdot i \) is an example of an alternating series. Each term alternates in sign due to the factor \((-1)^{i+1}\). This provides a pattern where terms repeatedly switch between being added and subtracted.

Such a series is essential in calculus and real analysis because it introduces the concept of convergence within changing sequences. To ensure that an alternating series converges, the magnitude of the terms should approach zero as the sequence progresses.

Summation Formula

The summation formula represents a compact way to add a sequence of numbers, often using sigma notation (\(\Sigma\)). This notation illustrates how to sum terms from a particular beginning index to an ending index. For example, the expression \( \sum_{i=1}^{5} i \) indicates that you sum all the integers from 1 to 5.

In our example, the summation formula \( \sum_{i=1}^{5} (-1)^{i+1} \cdot i \) illustrates the evaluation of an alternating series. Each term is calculated by substituting integer values from 1 to 5 into the formula part \((-1)^{i+1} \cdot i\).

Sigma notation is powerful for keeping expressions concise. It allows mathematicians to represent long sums without explicitly writing every single term. This notation is prevalent in calculus, algebra, and statistical formulas due to its ability to simplify the articulation of complex summative expressions.