Problem 5 - Calculus Volume 1

Question

Suppose that \(\sum_{i=1}^{100} a_{i}=15\) and \(\sum_{i=1}^{100} b_{i}=-12 .\) In the following exercises, compute the sums. $$\sum_{i=1}^{100}\left(a_{i}-b_{i}\right)$$

Step-by-Step Solution

Answer

The sum \( \sum_{i=1}^{100}(a_{i}-b_{i}) \) is 27.

Step 1: Understand the Problem

We are given two separate summations: \( \sum_{i=1}^{100} a_{i}=15 \) and \( \sum_{i=1}^{100} b_{i}=-12 \). We need to find \( \sum_{i=1}^{100} (a_{i} - b_{i}) \). This involves subtracting each \( b_{i} \) from the corresponding \( a_{i} \) and then summing these differences from \( i=1 \) to \( i=100 \).

Step 2: Apply the Sum Property

Understand that the sum of the differences \( \sum_{i=1}^{100} (a_{i} - b_{i}) \) can be rewritten by distributing the summation over the subtraction: \( \sum_{i=1}^{100} a_{i} - \sum_{i=1}^{100} b_{i} \). This is by applying the property of summations that \( \sum (a_i - b_i) = \sum a_i - \sum b_i \).

Step 3: Substitute the Given Values

Now, substitute the given values into the rewritten equation: \( \sum_{i=1}^{100} a_{i} - \sum_{i=1}^{100} b_{i} = 15 - (-12) \).

Step 4: Perform the Calculation

Calculate the result by subtracting: \( 15 - (-12) = 15 + 12 = 27 \). By simplifying, we find that the sum of the differences is 27.

Key Concepts

Summation Properties

Summation properties are fundamental tools in mathematics that make working with series of numbers more manageable. In this exercise, we specifically use the linearity property of summations. This property states that the summation of differences is equivalent to the difference of summations.
For example, if you have two sequences, represented by \(a_i\) and \(b_i\), across the same index range, i.e., from 1 to 100, the summation of their difference \((a_i - b_i)\) is simply:

• \(\sum_{i=1}^{100} (a_i - b_i) = \sum_{i=1}^{100} a_i - \sum_{i=1}^{100} b_i\)

This property is immensely useful when dealing with problems that require breaking down complex expressions into simpler components. It allows you to work with each part of a series independently, before combining the results. This simplification is what helps solve this problem efficiently.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations and expressions to simplify them or to find unknowns. In our problem, we need to find the sum of \((a_i - b_i)\) from 1 to 100. Given two individual summations, algebraic manipulation is used to transform our target equation into a form that's easier to calculate.

• We started with the equation \(\sum_{i=1}^{100} (a_i - b_i)\).

By applying summation properties, we rewrote it as \(\sum_{i=1}^{100} a_i - \sum_{i=1}^{100} b_i\). Once this transformation is done, the next step is simple arithmetic involving the given sums.
Substitute the known values for the sums: \(15 - (-12)\). Since subtracting a negative is the same as adding a positive, you get \(15 + 12\).
This is a straightforward example of utilizing both algebraic manipulation and arithmetic to solve expressions easily.

Calculus Problem Solving

Calculus problem solving often employs both summation properties and algebraic manipulation to solve problems effectively. While this exercise might not directly involve derivatives or integrals, the principles behind calculus—like focusing on incremental changes and summation—are demonstrated.

• Recognize how breaking down a problem using summation can simplify the overall structure.

Understanding how calculus integrates these properties helps in more advanced mathematical contexts. Consider this structured approach:

• Identify given values and aim—\(\sum_{i=1}^{100} (a_i - b_i)\).
• Apply summation rules to simplify.
• Substitute and calculate using known values.

This logical workflow is akin to solving calculus problems, where setting up the expression and simplifying with the right properties leads to a solution. Thus, mastering these techniques prepares you for tackling complex calculus scenarios.