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Chapter 5: Problem 2

Without sigma notation. Then evaluate them. $$\sum_{k=1}^{3} \frac{k-1}{k}$$

Short Answer

Expert verified
The sum is \( \frac{7}{6} \).

Step by step solution

01

Expand the Series without Sigma Notation

First, let's interpret the given sigma notation. The series is the sum of terms of the form \( \frac{k-1}{k} \) where \( k \) ranges from 1 to 3. This means we need to evaluate and sum the following terms: \( \frac{1-1}{1} \), \( \frac{2-1}{2} \), \( \frac{3-1}{3} \).
02

Compute Individual Terms

Next, compute each term individually:- For \( k = 1 \), \( \frac{1-1}{1} = \frac{0}{1} = 0 \).- For \( k = 2 \), \( \frac{2-1}{2} = \frac{1}{2} \).- For \( k = 3 \), \( \frac{3-1}{3} = \frac{2}{3} \).
03

Sum the Terms

Now, sum up the computed terms: \( 0 + \frac{1}{2} + \frac{2}{3} = \frac{0}{1} + \frac{1}{2} + \frac{2}{3} \).
04

Find a Common Denominator for Addition

To add the fractions, find a common denominator. The least common multiple of 2 and 3 is 6. Convert each fraction:- \( \frac{1}{2} = \frac{3}{6} \)- \( \frac{2}{3} = \frac{4}{6} \)Sum: \( 0 + \frac{3}{6} + \frac{4}{6} = \frac{3+4}{6} = \frac{7}{6} \)
05

Simplify the Result if Possible

The fraction \( \frac{7}{6} \) is already in its simplest form.

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Key Concepts

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

Sigma Notation
Sigma notation, denoted by the Greek letter \( \Sigma \), is a concise way to express the sum of a sequence of numbers. It's widely used in mathematics to simplify the representation of summation. In this particular exercise, sigma notation is used to describe the series \( \sum_{k=1}^{3} \frac{k-1}{k} \).

Here's a breakdown of the different parts:
  • The \( \Sigma \) symbol itself stands for 'sum.'
  • Below it, \( k=1 \) specifies the starting value of the index \( k \).
  • Above it, \( 3 \) indicates the ending value for \( k \).
  • The expression \( \frac{k-1}{k} \) is the formula for each term as \( k \) takes values from 1 to 3.
To solve the series, replace \( k \) in the expression \( \frac{k-1}{k} \) with each integer from 1 through 3, calculate each term, and then add them together.
Fraction Addition
Adding fractions is a fundamental process in arithmetic that requires combining fractions to produce a single result. For the series we are evaluating, the terms involve fractions that need to be summed:

First, compute each term separately:
  • For \( k = 1 \), the term is \( \frac{0}{1} = 0 \).
  • For \( k = 2 \), calculate \( \frac{1}{2} \).
  • For \( k = 3 \), it's \( \frac{2}{3} \).
Once these values are obtained, they need to be summed: \( 0 + \frac{1}{2} + \frac{2}{3} \). Fraction addition requires a common denominator, which is addressed in the next section.
Common Denominator
For adding fractions like \( \frac{1}{2} \) and \( \frac{2}{3} \), finding a common denominator is necessary. This means identifying the least common multiple (LCM) of the denominators involved, here 2 and 3.

Steps to find a common denominator:
  • The LCM of 2 and 3 is 6. This is the smallest number that both 2 and 3 divide into evenly.
  • Convert \( \frac{1}{2} \) to an equivalent fraction with the denominator 6 by finding \( \frac{1 \times 3}{2 \times 3} = \frac{3}{6} \).
  • Convert \( \frac{2}{3} \) to \( \frac{2 \times 2}{3 \times 2} = \frac{4}{6} \).
Now, sum the fractions with their newly found common denominator:\[\frac{3}{6} + \frac{4}{6} = \frac{7}{6}\], which results in a simplified fraction already in its simplest form.

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Most popular questions from this chapter

If \(f\) is a continuous function, find the value of the integral $$I=\int_{0}^{a} \frac{f(x) d x}{f(x)+f(a-x)}$$ by making the substitution \(u=a-x\) and adding the resulting integral to \(I\)

Use a CAS to perform the following steps: a. Plot the functions over the given interval. b. Partition the interval into \(n=100.200\), and 1000 subintervals of equal length, and evaluate the function at the midpoint of each subinterval. c. Compute the average value of the function values generated in part (b). d. Solve the equation \(f(x)=\) (average value) for \(x\) using the average value calculated in part (c) for the \(n=1000\) partitioning. $$f(x)=x \sin \frac{1}{x} \quad \text { on } \quad\left[\frac{\pi}{4}, \pi\right]$$

What values of \(a\) and \(b\) maximize the value of $$ \int_{a}^{b}\left(x-x^{2}\right) d x ? $$ (Hint: Where is the integrand positive?)

Show that if \(f\) is continuous, then $$\int_{0}^{1} f(x) d x=\int_{0}^{1} f(1-x) d x$$

Find \(d y / d x\). $$y=\int_{2^{x}}^{1} \sqrt[3]{t} d t$$
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