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Chapter 5: Problem 17

Express the sums in closed form. $$\sum_{k=1}^{n} \frac{3 k}{n}$$

Short Answer

Expert verified
\( \frac{3(n+1)}{2} \)

Step by step solution

01

Understanding the Summation

We are given the sum \( \sum_{k=1}^{n} \frac{3k}{n} \). This is a sum over the index \( k \) where each term in the sum is \( \frac{3k}{n} \).
02

Factor Out Constants

Notice that the fraction \( \frac{3}{n} \) is constant with respect to \( k \), so it can be factored out of the summation: \( \sum_{k=1}^{n} \frac{3k}{n} = \frac{3}{n} \sum_{k=1}^{n} k \).
03

Recognize the Arithmetic Series

The inner summation \( \sum_{k=1}^{n} k \) is a well-known arithmetic series with a closed form. The formula for the sum of the first \( n \) positive integers is \( \frac{n(n+1)}{2} \).
04

Apply the Formula for the Sum of Integers

Substitute the closed form of the arithmetic series \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \) into the expression to get: \( \frac{3}{n} \cdot \frac{n(n+1)}{2} \).
05

Simplify the Expression

Simplify the expression: \( \frac{3n(n+1)}{2n} = \frac{3(n+1)}{2} \). This is the closed form expression for the original summation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arithmetic Series
An arithmetic series is a sequence of numbers in which the difference between consecutive terms is constant. This difference is known as the common difference. For example, the series 2, 4, 6, 8,... is an arithmetic series with a common difference of 2.
An important characteristic of an arithmetic series is that it can be summed up quickly. If you have a series like the first \( n \) numbers \( 1, 2, 3, \ldots, n \), this forms an arithmetic series. Here, the common difference is 1. Understanding these basics is essential, as it sets the foundation for finding a closed form of the sum of its terms.
  • If you can recognize an arithmetic series, you can use a simple formula to find its sum.
  • An arithmetic series simplifies calculations by allowing quick and efficient summation.
Sum of Integers Formula
The sum of integers formula is a handy tool in solving series problems. Specifically, it looks at the sum of the first \( n \) integers, represented by the formula \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \).
This formula emerges because you're essentially adding the entire set of integers, which simplifies using sum formulas. When you lay out the numbers from 1 to \( n \), you can pair them from opposite ends of the list, each pair summing to \( n+1 \).
  • This formula is straightforward to apply and offers a quick path to the sum.
  • It helps in understanding other complex summations by breaking them into simpler arithmetic series.
When tackling problems, recognizing this pattern can lead to its swift application, like identifying a known formula in your toolbox.
Factor Out Constants
In the study of summations, one useful technique is to factor out constants. This means taking constants outside the summation symbol, focusing the calculation only on the variable term. When every term in a summation shares a constant multiplier, that constant can be placed outside
using the property \( \sum_{k=1}^{n} c \cdot a_k = c \cdot \sum_{k=1}^{n} a_k \).
This technique greatly simplifies the calculations, allowing focus on the more variable part of the equation. Factoring out constants can drastically reduce computational effort and make the algebra more manageable.
  • It's akin to simplifying a problem before you solve it, ensuring you're not bogged down by repetitive multiplication.
  • This is crucial when working alongside other techniques, such as using a known formula for the variable part.
In our problem, we used this strategy by factoring out \( \frac{3}{n} \), leaving us only to calculate the sum of integers.

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Most popular questions from this chapter

Electricity is supplied to homes in the form of alternating current, which means that the voltage has a sinusoidal waveform described by an equation of the form $$V=V_{p} \sin (2 \pi f t)$$ (see the accompanying figure). In this equation, \(V_{p}\) is called the peak voltage or amplitude of the current, \(f\) is called its frequency, and \(1 / f\) is called its period. The voltages \(V\) and \(V_{p}\) are measured in volts (V), the time \(t\) is measured in seconds (s), and the frequency is measured in hertz (Hz). (1 \(\mathrm{Hz}=1\) cycle per second; a cycle is the electrical term for one period of the waveform.) Most alternating-current voltmeters read what is called the rms or root -mean-square value of \(V .\) By definition, this is the square root of the average value of \(V^{2}\) over one period. (a) Show that $$V_{\mathrm{rms}}=\frac{V_{p}}{\sqrt{2}}$$ [Hint: Compute the average over the cycle from \(t=0\) to \(t=1 / f,\) and use the identity \(\sin ^{2} \theta=\frac{1}{2}(1-\cos 2 \theta)\) to help evaluate the integral. \(]\) (b) In the United States, electrical outlets supply alternating current with an rms voltage of \(120 \mathrm{V}\) at a frequency of \(60 \mathrm{Hz}\). What is the peak voltage at such an outlet? (Check your book to see figure)

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