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Chapter 5: Problem 6

Write each expression in sigma notation but do not evaluate. $$ 1+3+5+7+\dots+15 $$

Short Answer

Expert verified
\( \sum_{n=1}^{8} (2n - 1) \)

Step by step solution

01

Identify the Pattern

We begin by examining the sequence: 1, 3, 5, 7, ..., 15. We notice that these are consecutive odd numbers starting from 1. The first term (1) is when the sequence starts.
02

Determine the General Term

The sequence of odd numbers can be expressed generally as \(2n - 1\), where \(n\) starts from 1 and increases to the point where \(2n - 1 = 15\).
03

Calculate the Number of Terms

To find how many terms are in the sequence, set \(2n - 1 = 15\). Solving for \(n\), \(2n = 16\) which leads to \(n = 8\). So, there are 8 terms in the sequence.
04

Write in Sigma Notation

Now that we have the general term \(2n - 1\) and the number of terms \(n\) going from 1 to 8, we can express the sum in sigma notation: \( \sum_{n=1}^{8} (2n - 1) \).

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Key Concepts

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

Sequence of Odd Numbers
A sequence of odd numbers is a straightforward concept once you identify the pattern. Odd numbers are those integers that cannot be divided evenly by 2, meaning when divided by 2, they leave a remainder of 1.
In this exercise, you encountered the sequence 1, 3, 5, 7, ..., 15. Here you can see that each subsequent number increases by 2, making them consecutive odd numbers.
To recognize a sequence of odd numbers:
  • Start with the number 1.
  • Add 2 continually to get the next odd number.
  • Continue this pattern indefinitely.
Identifying such a sequence helps simplify working with numbers in various mathematical calculations, particularly summations.
General Term
The general term of a sequence is a formula that helps you find any term in the sequence without having to list all previous terms.
For a sequence of odd numbers, the general expression is simple yet powerful: \(2n - 1\). This formula allows you to generate any arbitrary odd number in the sequence by plugging in a different value of \(n\), where \(n\) is the term number.
Here's how it works:
  • For the 1st term, plug in \(n = 1\): \(2(1) - 1 = 1\)
  • For the 2nd term, plug in \(n = 2\): \(2(2) - 1 = 3\)
  • Continue the pattern: \(2n - 1\)
This general term makes calculations efficient and forms the basis for expressing the sequence in sigma notation, key for summing a sequence without calculating each individual term.
Number of Terms
Determining the number of terms in a sequence is critical to setting up expressions in sigma notation. To find how long a sequence stretches, use the general term to solve for \(n\).
In the odd number sequence leading to 15, you set \(2n - 1 = 15\) and solve for \(n\). This involves basic algebra:
  • Add 1 to both sides to get \(2n = 16\).
  • Divide both sides by 2 to find \(n = 8\).
Therefore, the sequence contains 8 terms.
Understanding this concept ensures you correctly define the limits for sigma notation, a powerful tool for representing and summing long sequences compactly.

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Most popular questions from this chapter

(a) Use a CAS to find the exact value of the integral $$ \int_{-\pi / 4}^{\pi / 4} \tan ^{4} x d x $$ (b) Confirm the exact value by hand calculation. [Hint: Use the identity \(\left.1+\tan ^{2} x=\sec ^{2} x .\right]\)

Evaluate the integrals by any method. $$ \int_{0}^{\pi / 6} \tan 2 \theta d \theta $$

(a) Find the limit $$\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{\sin (k \pi / n)}{n}$$ by evaluating an appropriate definite integral over the interval \([0,1] .\) (b) Check your answer to part (a) by evaluating the limit directly with a CAS.

(a) Over what open interval does the formula $$ F(x)=\int_{1}^{x} \frac{d t}{t} $$ represent an antiderivative of \(f(x)=1 / x ?\) (b) Find a point where the graph of \(F\) crosses the \(x\) -axis.

Writing A student objects that it is circular reasoning to make the definition $$ \ln x=\int_{1}^{x} \frac{1}{t} d t $$ since to evaluate the integral we need to know the value of \(\ln x\). Write a short paragraph that answers this student's objection.
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