Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that \(\sum_{i=1}^{10} a_{i}=40\) and \(\sum_{i=1}^{10} b_{i}=50 .\) Calculate each of the following. $$ \sum_{n=1}^{10}\left(3 a_{n}+2 b_{n}\right) $$

Step-by-Step Solution

Answer

The value is 220.

Step 1: Expand the Given Summation

We need to calculate \( \sum_{n=1}^{10} (3a_n + 2b_n) \). First, we can use the property of summation to separate \(3a_n\) and \(2b_n\) as follows:\[ \sum_{n=1}^{10} (3a_n + 2b_n) = \sum_{n=1}^{10} 3a_n + \sum_{n=1}^{10} 2b_n \]

Step 2: Factor Out the Constants

Next, factor out the constants from each part of the separated sums:\[ \sum_{n=1}^{10} 3a_n = 3 \sum_{n=1}^{10} a_n \] and \[ \sum_{n=1}^{10} 2b_n = 2 \sum_{n=1}^{10} b_n \]

Step 3: Substitute Given Values

Substitute the given values \( \sum_{i=1}^{10} a_i = 40 \) and \( \sum_{i=1}^{10} b_i = 50 \) into the expression:\[ 3 \sum_{n=1}^{10} a_n = 3 \times 40 = 120 \] and\[ 2 \sum_{n=1}^{10} b_n = 2 \times 50 = 100 \]

Step 4: Add the Results

Add the results from the calculations above to find the value of the original summation:\[ 120 + 100 = 220 \]

Key Concepts

Understanding Properties of Summation

The properties of summation are extremely helpful when dealing with arithmetic and calculus problems. They allow us to break down complex summations into simpler parts. This way, we can manage large equations with ease and clarity. One of these properties is linearity, which permits the sum of two separate sequences, as seen in our example:

• We are given \( \sum_{n=1}^{10} (3a_n + 2b_n) \).
• This expression can be split into \( \sum_{n=1}^{10} 3a_n + \sum_{n=1}^{10} 2b_n \).

By applying the linearity property, we simplify both arithmetic and algebraic manipulations. Moreover, summations can often be restructured, enabling the factoring out of constants and thus making further calculations more straightforward.

Handling Constant Factors in Summation

Dealing with constant factors in summation is straightforward, thanks to another key property of summation: the constant factor rule. This rule allows you to "factor out" constants from inside a summation.

• In our example, we separate \( 3 \) from \( 3a_n \) and \( 2 \) from \( 2b_n \), allowing us to rewrite the sums as \( 3 \sum_{n=1}^{10} a_n \) and \( 2 \sum_{n=1}^{10} b_n \).

This move is valuable because it allows you to multiply the sum of terms by a constant without changing the overall meaning of the sum. It makes broader calculations more manageable since you can first find the summation and then simply multiply by the constant. This strategy takes one big calculation and turns it into multiple simpler steps, which can be tackled independently.

Connecting Calculus and Summation

Understanding summation is fundamental in calculus, particularly when dealing with integration and differentiation. These processes can be viewed as the summation of infinitesimally small quantities—a concept central to calculus.

• Simplifying sums, as we've done with \( \sum_{n=1}^{10} (3a_n + 2b_n) \), helps develop intuition for more complex integration techniques.
• In calculus, the summation of a sequence can set the stage for understanding definite integrals, where limits of integration take the role of summation indices.

Moreover, recognizing and manipulating constant factors in summations is crucial in calculus applications, where constants regularly appear in problems involving rates of change and area calculations. Mastering these summation techniques brings clarity and ease into solving basic and advanced calculus problems.