Problem 18 - University Calculus: Early Transcendentals

Question

Suppose that \(\sum_{k=1}^{n} a_{k}=0\) and \(\sum_{k=1}^{n} b_{k}=1 .\) Find the values of a. \(\sum_{k=1}^{n} 8 a_{k}\) b. \(\sum_{k=1}^{n} 250 b_{k}\) c. \(\sum_{k=1}^{n}\left(a_{k}+1\right)\) d. \(\sum_{k=1}^{n}\left(b_{k}-1\right)\)

Step-by-Step Solution

Answer

a = 0; b = 250; c = n; d = 1 - n.

Step 1: Analyzing the First Expression

For part a, you are given \( \sum_{k=1}^{n} a_{k} = 0 \). You need to find \( \sum_{k=1}^{n} 8a_{k} \). Since \( 8 \) is a constant, you can factor it out from the sum. Thus, \( \sum_{k=1}^{n} 8a_{k} = 8 \sum_{k=1}^{n} a_{k} \). Since \( \sum_{k=1}^{n} a_{k} = 0 \), \( \sum_{k=1}^{n} 8a_{k} = 8 \times 0 = 0 \).

Step 2: Solving the Second Expression

For part b, you are given \( \sum_{k=1}^{n} b_{k} = 1 \). You need to find \( \sum_{k=1}^{n} 250 b_{k} \). With the constant \( 250 \) being factorable, you get \( \sum_{k=1}^{n} 250b_{k} = 250 \times \sum_{k=1}^{n} b_{k} = 250 \times 1 = 250 \).

Step 3: Evaluating the Third Expression

In part c, we consider \( \sum_{k=1}^{n} (a_{k} + 1) \). This expression can be split into \( \sum_{k=1}^{n} a_{k} + \sum_{k=1}^{n} 1 \). We know \( \sum_{k=1}^{n} a_{k} = 0 \). \( \sum_{k=1}^{n} 1 \) represents adding \( 1 \) for each term from 1 to \( n \), which is simply \( n \). So \( \sum_{k=1}^{n} (a_{k} + 1) = 0 + n = n \).

Step 4: Completing the Fourth Expression

Finally, for part d, analyze \( \sum_{k=1}^{n} (b_{k} - 1) \). This can be split as \( \sum_{k=1}^{n} b_{k} - \sum_{k=1}^{n} 1 \). Given that \( \sum_{k=1}^{n} b_{k} = 1 \) and \( \sum_{k=1}^{n} 1 = n \), the result is \( 1 - n \). Therefore, \( \sum_{k=1}^{n} (b_{k} - 1) = 1 - n \).

Key Concepts

Sum of series

Summation is a core concept in calculus often used to add up sequences of numbers. When dealing with series, the notation \( \sum_{k=1}^{n} a_{k} \) is used to represent the sum of all terms \( a_{k} \) from \( k=1 \) to \( k=n \). Understanding this notation is important because it allows us to simplify and calculate lengthy sums efficiently. For example, in our problem, we have two different series: \( \sum_{k=1}^{n} a_{k} = 0 \) and \( \sum_{k=1}^{n} b_{k} = 1 \). This gives us the totals of these series without having to actually add each individual term, making summation a powerful tool in calculus. Summation notation helps encapsulate complex series simplistically, allowing mathematicians to work with them analytically instead of computationally.

Constant factor in summation

When dealing with summations, it's important to recognize how constants can be factored out. This property can greatly reduce the complexity of the expression.If you have a constant \( c \) and a series \( \sum_{k=1}^{n} a_{k} \), the constant can be multiplied after summing the series: \[ \sum_{k=1}^{n} c \cdot a_{k} = c \cdot \sum_{k=1}^{n} a_{k} \]In practical terms, this means if the summation value is known, instead of multiplying each term by the constant, you can factor it out and multiply it at the end.For instance, with \( \sum_{k=1}^{n} a_{k} = 0 \), calculating \( \sum_{k=1}^{n} 8a_{k} \) becomes: \[ 8 \times \sum_{k=1}^{n} a_{k} = 8 \times 0 = 0 \]Thus, a constant factor in summation simplifies calculations and provides quick insights into the series.

Splitting summations

The ability to split summations lets us break down complex series into simpler components, making them easier to manage and solve. When you encounter a series such as \( \sum_{k=1}^{n} (a_{k} + b_{k}) \), you can separate it into two distinct series:\[ \sum_{k=1}^{n} a_{k} + \sum_{k=1}^{n} b_{k} \]In our problem, this principle is used to solve examples like \( \sum_{k=1}^{n} (a_{k} + 1) \). Here we split it to:\[ \sum_{k=1}^{n} a_{k} + \sum_{k=1}^{n} 1 \]Since \( \sum_{k=1}^{n} a_{k} = 0 \) and \( \sum_{k=1}^{n} 1 = n \) (as adding 1 repeatedly for each term just multiplies 1 by \( n \)), the result is simple:\[ 0 + n = n \] This property of splitting summations is essential for simplifying and analyzing series quickly.

Properties of summations

Properties of summations are rules that govern how we can manipulate and work with summation notation. These rules make it easier to handle various operations without excessive calculations. Some key properties are:

• Additivity: \( \sum_{k=1}^{n} (a_{k} + b_{k}) = \sum_{k=1}^{n} a_{k} + \sum_{k=1}^{n} b_{k} \)
• Subtraction: \( \sum_{k=1}^{n} (a_{k} - b_{k}) = \sum_{k=1}^{n} a_{k} - \sum_{k=1}^{n} b_{k} \)
• Constant Multiplication: \( \sum_{k=1}^{n} c \cdot a_{k} = c \cdot \sum_{k=1}^{n} a_{k} \)

In the original exercise, these properties were utilized effectively. For example, when calculating \( \sum_{k=1}^{n} (b_{k} - 1) \), we see the property of subtraction being used:\[ \sum_{k=1}^{n} b_{k} - \sum_{k=1}^{n} 1 \]Given that \( \sum_{k=1}^{n} b_{k} = 1 \) and \( \sum_{k=1}^{n} 1 = n \), the result is \( 1 - n \).Understanding these properties aids in simplifying expressions and solving problems more efficiently.