Problem 29 - Calculus for Biology and Medicine

Question

Use the algebraic rules for sums to evaluate each sum. Recall that $$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$$ and $$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$$ $$ \sum_{k=1}^{10}(-1)^{k} $$

Step-by-Step Solution

Answer

The sum is 0.

Step 1: Understand the Series

The series given is \(\sum_{k=1}^{10}(-1)^{k}\). This sum alternates signs based on the exponent of \((-1)\), producing 1 for odd \(k\) and -1 for even \(k\).

Step 2: Identify the Pattern

List the first few terms to see the pattern: \((-1)^1 = -1\), \((-1)^2 = 1\), \((-1)^3 = -1\), and so on. Each pair of consecutive terms, such as \( (-1) + 1 \), sums to zero.

Step 3: Count the Complete Pairs

Since we have 10 terms (from 1 to 10 inclusive), there are 5 complete pairs: \((-1)^1 + (-1)^2, (-1)^3 + (-1)^4, ..., (-1)^9 + (-1)^{10}\).

Step 4: Summation of the Series

Each pair of terms \((-1) + 1\) sums to zero. Therefore, all 5 pairs together sum to zero: \(0 + 0 + 0 + 0 + 0 = 0\).

Key Concepts

Series Evaluation

Series evaluation involves finding the sum of a sequence of numbers following a certain pattern. In our case with the exercise, the sum is defined as \( \sum_{k=1}^{10}(-1)^{k} \). Understanding the makeup of this series is key to accurately evaluating it. The sum is straightforward to calculate once you grasp the behavior of its terms.

• The series is composed of terms like \((-1)^{k}\), where the outcome is based on whether \(k\) is odd or even.
• Odd \(k\) values yield \(-1\), while even \(k\) values yield \(+1\).

So, when evaluating this series, you quickly notice there is a repeating pattern that simplifies the calculation. The fundamental algebraic rule for sums tells us that completing the pairs simplifies the formulation and ultimately the solution. Therefore, leveraging an observation or a rule helps effectively in evaluating complicated series.

Alternating Series

An alternating series is characterized by terms that switch signs as the series progresses. The given series, \( \sum_{k=1}^{10} (-1)^k \), is a classic example of an alternating series.

• For odd \(k\), the terms will be negative, like \((-1)^1 = -1\).
• For even \(k\), the terms will be positive, like \((-1)^2 = 1\).

The alteration of signs makes a significant impact, as it often simplifies computations through cancellation. In the problem's scenario, these alternating signs naturally create pairs that sum up to zero. Notice how this alternating nature determines the behavior of the series drastically. When the series reaches a total count of terms that result in pairs, all resulting pairs neutralize each other.

Pattern Recognition

Pattern recognition is essential for handling series that might initially seem complex. By analyzing the first few terms of the series \((-1)^1, (-1)^2, (-1)^3, \) you notice a pattern of \(-1, +1, -1, +1,...\).

• Each consecutive pair such as \((-1) + 1\) equates to zero.
• Understanding how terms recur helps quickly summarizing the result of the series.

We recognize this pattern and use it to foresee its behavior over a longer sequence of terms. Especially for an alternating series, identifying this pattern is crucial. When the pattern of sign change is clear, the computation becomes much more manageable.This pattern-based simplification is what makes the evaluation process both intuitive and efficient.