Problem 17 - Thomas Calculus

Question

Suppose that \(\sum_{k=1}^{n} a_{k}=-5\) and \(\sum_{k=1}^{n} b_{k}=6 .\) Find the values of $$ \begin{array}{ll}{\text { a. } \sum_{k=1}^{n} 3 a_{k}} & {\text { b. } \sum_{k=1}^{n} \frac{b_{k}}{6}} & {\text { c. } \sum_{k=1}^{n}\left(a_{k}+b_{k}\right)} \\ {\text { d. } \sum_{k=1}^{n}\left(a_{k}-b_{k}\right)} & {\text { e. } \sum_{k=1}^{n}\left(b_{k}-2 a_{k}\right)}\end{array} $$

Step-by-Step Solution

Answer

a. -15, b. 1, c. 1, d. -11, e. 16.

Step 1: Find \( \sum_{k=1}^{n} 3 a_{k} \)

We are given \( \sum_{k=1}^{n} a_{k} = -5 \). To find \( \sum_{k=1}^{n} 3 a_{k} \), we simply multiply the entire given sum by 3 since it is a linear operation. Thus, \( \sum_{k=1}^{n} 3 a_{k} = 3 \times (-5) = -15 \).

Step 2: Find \( \sum_{k=1}^{n} \frac{b_{k}}{6} \)

Given \( \sum_{k=1}^{n} b_{k} = 6 \), to find \( \sum_{k=1}^{n} \frac{b_{k}}{6} \), divide the sum by 6. Thus, \( \sum_{k=1}^{n} \frac{b_{k}}{6} = \frac{6}{6} = 1 \).

Step 3: Find \( \sum_{k=1}^{n} (a_{k}+b_{k}) \)

Using the distributive property of sums, \( \sum_{k=1}^{n} (a_{k} + b_{k}) = \sum_{k=1}^{n} a_{k} + \sum_{k=1}^{n} b_{k} \). Plug in the given sums: \( \sum_{k=1}^{n} (a_{k} + b_{k}) = -5 + 6 = 1 \).

Step 4: Find \( \sum_{k=1}^{n} (a_{k} - b_{k}) \)

Similarly, \( \sum_{k=1}^{n} (a_{k} - b_{k}) = \sum_{k=1}^{n} a_{k} - \sum_{k=1}^{n} b_{k} \). Using the given values, \( \sum_{k=1}^{n} (a_{k} - b_{k}) = -5 - 6 = -11 \).

Step 5: Find \( \sum_{k=1}^{n} (b_{k} - 2a_{k}) \)

Here, \( \sum_{k=1}^{n} (b_{k} - 2a_{k}) = \sum_{k=1}^{n} b_{k} - 2 \cdot \sum_{k=1}^{n} a_{k} \). Substitute the given sums: \( \sum_{k=1}^{n} (b_{k} - 2a_{k}) = 6 - 2 \times (-5) = 6 + 10 = 16 \).

Key Concepts

Linear Operations

Linear operations are mathematical manipulations involving the addition, subtraction, or multiplication of any equation or expression. They are called "linear" because the operations are performed on linear forms of numbers or variables. In the context of series summation, linear operations allow us to simplify expressions efficiently.
For example, when you encounter the summation \( \sum_{k=1}^{n} 3a_k \), this is simply a linear scaling of the original summation \( \sum_{k=1}^{n} a_k \). By using a linear operation—namely multiplication by 3—you can directly compute the result without calculating each term individually. If \( \sum_{k=1}^{n} a_k = -5 \), then multiplying the entire sum by 3 gives \( 3 \times (-5) = -15 \). This shows how linear operations can efficiently adjust the terms within a series.
Using linear operations simplifies calculations and you can follow similar steps for any scalar multiplication or division. Just remember to apply the scalar to the entire summation.

Distributive Property

The distributive property is a fundamental algebraic principle used to simplify expressions and compute summations more easily. It states that multiplying a sum by a number is equivalent to multiplying each addend individually and then summing the results.
In series summation, this property is very useful. Consider the expression \( \sum_{k=1}^{n} (a_k + b_k) \). According to the distributive property, we can break this down as follows:

• Distribute: \( (a_k + b_k) = a_k + b_k \)
• Sum individually: \( \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k \)

When the exercise gives us \( \sum_{k=1}^{n} a_k = -5 \) and \( \sum_{k=1}^{n} b_k = 6 \), we can directly compute \( \sum_{k=1}^{n} (a_k + b_k) = -5 + 6 = 1 \).
This property enables operations on sums that make complex problems more approachable and provides a clear path to finding solutions.

Algebraic Manipulation

Algebraic manipulation involves rearranging and rewriting expressions to make them simpler to work with or to extract useful information. This is common when dealing with summations as it allows us to transform the original problems into more straightforward tasks.
Take the expression \( \sum_{k=1}^{n} (b_k - 2a_k) \). By understanding the underlying structure with algebraic manipulation, we can separate it into:

• Separate the terms: \( \sum_{k=1}^{n} b_k - 2 \sum_{k=1}^{n} a_k \)
• Substitute known values: Given \( \sum_{k=1}^{n} b_k = 6 \) and \( \sum_{k=1}^{n} a_k = -5 \), we substitute to get \( 6 - 2(-5) \)

This manipulation gives us \( 6 + 10 = 16 \) indicating how algebraic rearrangement can simplify problem-solving processes.
Effective algebraic manipulation not only aids in achieving correct answers but also helps to deepen the understanding of how mathematical concepts interconnect.