Chapter 5: Problem 20
Express the sums in closed form. $$ \sum_{k=1}^{n}\left(\frac{5}{n}-\frac{2 k}{n}\right) $$
Short Answer
Expert verified
The closed form of the sum is \( 4 - n \).
Step by step solution
01
Rewrite the Sum
Express the given sum as two separate sums. The expression inside the sum is \( \frac{5}{n} - \frac{2k}{n} \). This can be rewritten as \( \sum_{k=1}^{n} \frac{5}{n} - \sum_{k=1}^{n} \frac{2k}{n} \).
02
Evaluate the First Sum
The first sum \( \sum_{k=1}^{n} \frac{5}{n} \) is a constant sum. Since the term \( \frac{5}{n} \) is constant, it can be factored out: \( \frac{5}{n} \sum_{k=1}^{n} 1 = \frac{5}{n} \cdot n = 5 \).
03
Evaluate the Second Sum
The second sum is \( \sum_{k=1}^{n} \frac{2k}{n} \). Factor out \( \frac{2}{n} \) from the sum: \( \frac{2}{n} \sum_{k=1}^{n} k \). The sum of the first \( n \) positive integers is \( \frac{n(n+1)}{2} \). Thus, \( \frac{2}{n} \sum_{k=1}^{n} k = \frac{2}{n} \cdot \frac{n(n+1)}{2} = n+1 \).
04
Combine the Results
Combine the results of the two sums from Step 2 and Step 3: \( 5 - (n+1) \). Simplify to get \( 4 - n \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
closed form
In mathematics, expressing a function or series in closed form means representing it with a finite number of operations. These operations might involve arithmetic, exponentials, or trigonometric functions. Closed forms provide a simplified and easy-to-understand result.
In the exercise, the goal is to express the sum \( \sum_{k=1}^{n}\left(\frac{5}{n}-\frac{2 k}{n}\right) \) in closed form. Through a series of algebraic manipulations, we derived that the closed-form solution is \( 4 - n \). This form offers a straightforward way to evaluate the sum without needing further calculations or iterations over the series. Closed-form solutions are especially useful because they reveal direct insights into the behavior and value of the expression as a whole.
In the exercise, the goal is to express the sum \( \sum_{k=1}^{n}\left(\frac{5}{n}-\frac{2 k}{n}\right) \) in closed form. Through a series of algebraic manipulations, we derived that the closed-form solution is \( 4 - n \). This form offers a straightforward way to evaluate the sum without needing further calculations or iterations over the series. Closed-form solutions are especially useful because they reveal direct insights into the behavior and value of the expression as a whole.
sequences and series
Sequences and series are fundamental concepts in mathematics. A sequence is an ordered list of numbers, while a series is the sum of the terms of a sequence. An understanding of these allows for deeper insights into how different numeric patterns can be summed or transformed.
In the context of the original exercise, we have a series composed of two separate sequences:
In the context of the original exercise, we have a series composed of two separate sequences:
- The first is a constant sequence \( \frac{5}{n} \) repeated \( n \) times.
- The second is a linear sequence \( \frac{2k}{n} \), where \( k \) ranges from 1 to \( n \).
sum of integers
The sum of integers is a common concept in mathematics that underpins many series evaluations. The sum of the first \( n \) natural numbers (1+2+3+...+n) can be calculated using the formula \( \frac{n(n+1)}{2} \).
In our problem, the second sum requires using this formula. The expression \( \sum_{k=1}^{n} k \) can be directly replaced by \( \frac{n(n+1)}{2} \). This formula is derived from the properties of arithmetic progressions and provides a quick way to find the total sum of sequential integers without computing each one individually.
In our problem, the second sum requires using this formula. The expression \( \sum_{k=1}^{n} k \) can be directly replaced by \( \frac{n(n+1)}{2} \). This formula is derived from the properties of arithmetic progressions and provides a quick way to find the total sum of sequential integers without computing each one individually.
- Know that for any positive integer \( n \), you can use the formula for the sum of integers to simplify calculations.
- It allows for a quick transition from a step-by-step addition to a single multiplication and division operation added to your existing algebraic structure.