/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Gives a formula for the \(n\) th... [FREE SOLUTION] | 91影视

91影视

Gives a formula for the \(n\) th term \(a_{n}\) of a sequence \(\left\\{a_{n}\right\\} .\) Find the values of \(a_{1}, a_{2}, a_{3},\) and \(a_{4}\). \(a_{n}=\frac{(-1)^{n+1}}{2 n-1}\)

Short Answer

Expert verified
\(a_1 = 1\), \(a_2 = -\frac{1}{3}\), \(a_3 = \frac{1}{5}\), \(a_4 = -\frac{1}{7}\).

Step by step solution

01

Understand the Formula

The formula given is \(a_{n} = \frac{(-1)^{n+1}}{2n-1}\). It consists of a numerator \((-1)^{n+1}\), which gives alternating signs depending on whether \(n\) is odd or even, and a denominator \(2n-1\). This hints at a sequence that alternates in sign.
02

Calculate a鈧

Substitute \(n = 1\) into the formula. This gives:\[a_1 = \frac{(-1)^{1+1}}{2 \times 1 - 1} = \frac{1}{1} = 1\]
03

Calculate a鈧

Substitute \(n = 2\) into the formula. This gives:\[a_2 = \frac{(-1)^{2+1}}{2 \times 2 - 1} = \frac{-1}{3}\]
04

Calculate a鈧

Substitute \(n = 3\) into the formula. This gives:\[a_3 = \frac{(-1)^{3+1}}{2 \times 3 - 1} = \frac{1}{5}\]
05

Calculate a鈧

Substitute \(n = 4\) into the formula. This gives:\[a_4 = \frac{(-1)^{4+1}}{2 \times 4 - 1} = \frac{-1}{7}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Alternating Signs
In sequences, alternating signs are used to create a pattern where the terms switch between positive and negative values. This sequence behavior is captured in the numerator of the formula, particularly by the expression \[(-1)^{n+1}\]Here:
  • If \(n\) is odd (like \(n = 1, 3, 5, \) etc.), \((-1)^{n+1}\) becomes \((-1)^2 = 1\), which is positive.
  • If \(n\) is even (like \(n = 2, 4, 6, \) etc.), \((-1)^{n+1}\) becomes \((-1)^3 = -1\), which is negative.
This technique of using powers of -1 is a common method in mathematics to alternate signs in sequences and series. By understanding this, you can predict the sign of any term in the sequence.
Nth Term
The sequence is defined by a general formula that generates its terms depending on the position \(n\). Here the formula given is:\[a_{n} = \frac{(-1)^{n+1}}{2n-1}\]The term "nth term" refers to the formula's ability to derive any term in the sequence when substituting different values of \(n\). This concept is useful when you want to calculate terms without having to sequentially compute each preceding one.Each \(n\) corresponds to one term in the sequence: \(a_1\), \(a_2\), \(a_3\), and so on. By plugging in values of \(n\), you can obtain corresponding terms of the sequence efficiently:
  • \(a_1\) when \(n = 1\)
  • \(a_2\) when \(n = 2\)
  • \(a_3\) when \(n = 3\)
  • \(a_4\) when \(n = 4\)
This is the fundamental principle behind any sequence: to produce terms methodically through the nth term formula.
Denominator
The denominator in a sequence formula plays a crucial role in determining the values of the sequence's terms. In our given sequence, it is defined as:\[2n-1\]This expression helps to ensure that the terms have different magnitudes, as each term is calculated by dividing the alternating sign by this value. As \(n\) increases, the denominator also increases, affecting the term's magnitude.Here's how it works:
  • When \(n = 1\), the denominator is \(2 \times 1 - 1 = 1\).
  • When \(n = 2\), it becomes \(2 \times 2 - 1 = 3\).
  • When \(n = 3\), it changes to \(2 \times 3 - 1 = 5\).
  • Continuing this pattern, \(n = 4\) yields a denominator of \(2 \times 4 - 1 = 7\).
The denominator affects the size of each term while the sign is governed by the numerator. Understanding both aspects is important for fully grasping how each sequence term is derived.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

a. Show that if two power series \(\Sigma_{n=0}^{\infty} a_{n} x^{n}\) and \(\Sigma_{n=0}^{\infty} b_{n} x^{n}\) are convergent and equal for all values of \(x\) in an open interval \((-c, c),\) then \(a_{n}=b_{n}\) for every \(n\) \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} b_{n} x^{n} .\) Differentiate term by term to show that \(a_{n}\) and \(b_{n}\) both equal \(f^{(n)}(0) /(n !)\) b. Show that if \(\Sigma_{n=0}^{\infty} a_{n} x^{n}=0\) for all \(x\) in an open interval \((-c, c),\) then \(a_{n}=0\) for every \(n\)

Assume that each sequence converges and find its limit. \(a_{1}=-1, \quad a_{n+1}=\frac{a_{n}+6}{a_{n}+2}\)

Which of the sequences converge, and which diverge? Give reasons for your answers. The first term of a sequence is \(x_{1}=\cos (1) .\) The next terms are \(x_{2}=x_{1}\) or \(\cos (2),\) whichever is larger; and \(x_{3}=x_{2}\) or \(\cos (3)\) whichever is larger (farther to the right). In general, $$x_{n+1}=\max \left\\{x_{n}, \cos (n+1)\right\\}$$.

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L\) ? b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) \(a_{1}=1, \quad a_{n+1}=a_{n}+\frac{1}{5^{n}}\)

If \(\Sigma a_{n}\) converges and \(\Sigma b_{n}\) diverges, can anything be said about their term-by-term sum \(\Sigma\left(a_{n}+b_{n}\right) ?\) Give reasons for your answer.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.