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Chapter 5: Problem 7

Write each expression in sigma notation but do not evaluate. $$1-3+5-7+9-11$$

Short Answer

Expert verified
The expression is \(\sum_{n=1}^{6} (-1)^{n+1} (2n-1)\).

Step by step solution

01

Identify the Pattern

Notice that the expression consists of alternating positive and negative odd numbers starting from 1. This suggests a pattern that can be captured using arithmetic sequence or series.
02

Define the General Term

Since the series alternates signs, we can express this using the general term for odd numbers: \(a_n = 2n - 1\). To alternate the signs, every other term is negative, so the general term becomes \((-1)^{n+1}(2n - 1)\).
03

Determine the Range of Summation

Count the total number of terms in the series. There are 6 terms, implying \(n\) goes from 1 to 6.
04

Write in Sigma Notation

Combine the general term and the range of summation to write the expression in sigma notation: \(\sum_{n=1}^{6} (-1)^{n+1} (2n-1)\).

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Key Concepts

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

Arithmetic Sequence
An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. This difference is known as the
  • common difference
  • or the step of the sequence.

In our problem, the sequence of numbers 1, 3, 5, 7, 9, 11 are increasing by 2 each time. Thus, the common difference, in this case, is 2. For arithmetic sequences, the general form of any term can be expressed as:

\[ a_n = a_1 + (n-1) imes d \]where:
  • \( a_n \) is the nth term of the sequence,
  • \( a_1 \) is the first term,
  • \( n \) is the term number,
  • \( d \) is the common difference.

Understanding arithmetic sequences helps identify the pattern in numbers and is crucial when we delve into more complex sequences and series.
Alternating Series
An alternating series is one where the terms change signs between positive and negative throughout the sequence. This is evident in our sequence: 1, -3, 5, -7, 9, -11.

Such sequences are significant because they oscillate around a particular range, unlike non-alternating sequences which grow in a single direction. To express alternating signs in a series, we often use \((-1)^{n+1}\) or \((-1)^{n}\).
  • When multiplied by the constant term, it flips the term between positive and negative values.
  • This results in a precise pattern of alternating signs.
This characteristic is used to derive the general term of the sequence and aids in capturing the essence of the sequence in sigma notation.
General Term in a Series
The general term in a series helps to represent each term in a consistent mathematical expression. For our sequence, the general term not only captures the arithmetic nature of the numbers but also integrates the alternating signs.

In mathematical terms, the general expression is:\[(-1)^{n+1} (2n - 1)\]
  • The \((-1)^{n+1}\) ensures that the sign alternates with each term.
  • The \(2n - 1\) represents the arithmetic pattern of the sequence, giving the odd numbers starting from 1.

The general term is a key component when expressing a sequence in sigma notation, as it succinctly captures both the numeric pattern and the sign alternation. By seeing the sequence through the lens of its general term, it becomes easier to manipulate and analyze in algebraic settings.

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Most popular questions from this chapter

Let $$S=\sum_{k=0}^{n} a r^{k}$$ Show that \(S-r S=a-a r^{n+1}\) and hence that $$ \sum_{k=0}^{n} a r^{k}=\frac{a-a r^{n+1}}{1-r} \quad(r \neq 1) $$ (A sum of this form is called a geometric sum.)

A function \(f(x)\) is defined piecewise on an interval. In these exercises: (a) Use Theorem 5.5 .5 to find the integral of \(f(x)\) over the interval. (b) Find an anti-derivative of \(f(x)\) on the interval. (c) Use parts (a) and (b) to verify Part 1 of the Fundamental Theorem of Calculus. $$f(x)=\left\\{\begin{array}{ll} x, & 0 \leq x \leq 1 \\ x^{2}, & 1

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Write a paragraph that explains informally what it means for a function to be "integrable."

(a) Evaluate the integral \(\int \sin x \cos x \, d x\) by two methods: first by letting \(u=\sin x,\) and then by letting \(u=\cos x\) (b) Explain why the two apparently different answers obtained in part (a) are really equivalent.
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