Problem 17 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Sum of Scaled Values

To solve for \( \sum_{k=1}^{n} 3a_{k} \), use the property that a constant can be factored out of the summation: \( \sum_{k=1}^{n} 3a_{k} = 3 \sum_{k=1}^{n} a_{k} \). Given that \( \sum_{k=1}^{n} a_{k} = -5 \), we substitute this into our equation: \( 3 \times (-5) = -15 \). Thus, \( \sum_{k=1}^{n} 3a_{k} = -15 \).

Step 2: Sum of Fraction of Values

For \( \sum_{k=1}^{n} \frac{b_{k}}{6} \), factor \( \frac{1}{6} \) out of the summation: \( \sum_{k=1}^{n} \frac{b_{k}}{6} = \frac{1}{6} \sum_{k=1}^{n} b_{k} \). Given \( \sum_{k=1}^{n} b_{k} = 6 \), substitute to get: \( \frac{1}{6} \times 6 = 1 \). Thus, \( \sum_{k=1}^{n} \frac{b_{k}}{6} = 1 \).

Step 3: Sum of Combined Values

For \( \sum_{k=1}^{n}(a_{k} + b_{k}) \), use the property that summation can be distributed: \( \sum_{k=1}^{n}(a_{k} + b_{k}) = \sum_{k=1}^{n} a_{k} + \sum_{k=1}^{n} b_{k} \). With \( \sum_{k=1}^{n} a_{k} = -5 \) and \( \sum_{k=1}^{n} b_{k} = 6 \), we get: \( -5 + 6 = 1 \). Thus, \( \sum_{k=1}^{n}(a_{k} + b_{k}) = 1 \).

Step 4: Sum of Differences

For \( \sum_{k=1}^{n}(a_{k} - b_{k}) \), apply distribution: \( \sum_{k=1}^{n}(a_{k} - b_{k}) = \sum_{k=1}^{n} a_{k} - \sum_{k=1}^{n} b_{k} \). With \( \sum_{k=1}^{n} a_{k} = -5 \) and \( \sum_{k=1}^{n} b_{k} = 6 \), we have: \( -5 - 6 = -11 \). Thus, \( \sum_{k=1}^{n}(a_{k} - b_{k}) = -11 \).

Step 5: Sum of Complex Differences

For \( \sum_{k=1}^{n}(b_{k} - 2a_{k}) \), distribute summation: \( \sum_{k=1}^{n} b_{k} - 2 \sum_{k=1}^{n} a_{k} \). Substituting known sums, \( 6 - 2 \times (-5) = 6 + 10 = 16 \). Therefore, \( \sum_{k=1}^{n}(b_{k} - 2a_{k}) = 16 \).

Key Concepts

Finite series

A finite series is a mathematical concept where we sum a specific number of terms in a sequence. In our exercise, concepts like \(\sum_{k=1}^{n} a_k = -5\) and \(\sum_{k=1}^{n} b_k = 6\) show finite series by indicating that we are summing from term 1 to term \(n\).

Finite series are helpful because they allow us to find quick results by adding numbers up. This is different from infinite series, which would go on forever.

• The sequence ends after a specific number of terms \(n\).
• Used in both arithmetic sequences where each term adds a constant, and geometric sequences where each term is a multiple of the previous one.

In the exercise provided, you see the clear finite limits of the summations with specified \(n\), and the result of each finite series is then used as a basis for further calculations.

Arithmetic operations

Arithmetic operations are basic computations involving addition, subtraction, multiplication, and division. These operations are applied extensively in calculus, especially when working with series, to simplify or transform expressions.

For example, when finding \(\sum_{k=1}^{n} 3a_k\), your arithmetic operation involves scaling each term by 3.

Scaling in Series

Scaling means multiplying each term of a series by a constant, which is straightforward because you multiply the result of the series by that constant.

• Multiply or divide the entire series by a constant.
• Distributing constants over a summation simplifies evaluation.

Similarly, for \(\sum_{k=1}^{n} \frac{b_k}{6}\), each term is divided by 6. The process involves pulling the constant (\(\frac{1}{6}\)) out, multiplying it by the result of the summation.

These operations transform the series into a simpler form, making calculations manageable.

Summation properties

The properties of summation provide essential rules that allow mathematical expressions involving series to be simplified or manipulated.

1. Distribuion: Distributing summation over addition or subtraction is a core property. It helps in solving expressions like \(\sum_{k=1}^{n}(a_k + b_k)\), where you separate the summations: \(\sum_{k=1}^{n}a_k + \sum_{k=1}^{n}b_k\).

2. Constant Factor: Pulling out constants is another property where a constant multiplying each term in the summation can be extracted and then multiplied with the total. For instance, \(3\sum_{k=1}^{n}a_k = 3(-5)\).

Understanding these properties helps in calculating sums efficiently, as you reassign and recombine terms into simpler units. This is vital in unraveling multi-component expressions into more manageable parts.

Sequences and series

Sequences are ordered lists of numbers, while series are sums of sequences.

Every element in a sequence is associated with a position called the index. For example, in the sequence \(a_k\), the index \(k\) denotes its position.

• The relationship between terms can define the sequence, like arithmetic sequences where each term equals the previous plus a constant.
• In geometric sequences, each term equals the previous term multiplied by a constant factor.

When transforming a sequence into a series, you add terms from the first to the specified last. In our exercise, the sequences \(a_k\) and \(b_k\) become series, which are then summed from 1 to \(n\). This process of summing gives insight into patterns and relationships between terms, providing the means to analyze and solve real-world problems using simple or compound relationships.