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Chapter 12: Problem 8

Find the first four terms and the 100th term of the sequence. \(a_{n}=(-1)^{n+1} \frac{n}{n+1}\)

Short Answer

Expert verified
First four terms: \(\frac{1}{2}, -\frac{2}{3}, \frac{3}{4}, -\frac{4}{5}\). 100th term: \(-\frac{100}{101}\).

Step by step solution

01

Understanding the Formula for Terms of the Sequence

The formula given for the sequence is \(a_n = (-1)^{n+1} \frac{n}{n+1}\). The sequence is defined using a general term formula where \(n\) denotes the position of the term.
02

Calculating the First Term

To find the first term \(a_1\), substitute \(n = 1\) into the formula. Thus, \(a_1 = (-1)^{1+1} \frac{1}{1+1} = (-1)^{2} \frac{1}{2} = 1 \times \frac{1}{2} = \frac{1}{2}\).
03

Calculating the Second Term

For the second term \(a_2\), substitute \(n = 2\) into the formula. Thus, \(a_2 = (-1)^{2+1} \frac{2}{2+1} = (-1)^{3} \frac{2}{3} = -1 \times \frac{2}{3} = -\frac{2}{3}\).
04

Calculating the Third Term

For the third term \(a_3\), substitute \(n = 3\) into the formula. Thus, \(a_3 = (-1)^{3+1} \frac{3}{3+1} = (-1)^{4} \frac{3}{4} = 1 \times \frac{3}{4} = \frac{3}{4}\).
05

Calculating the Fourth Term

For the fourth term \(a_4\), substitute \(n = 4\) into the formula. Thus, \(a_4 = (-1)^{4+1} \frac{4}{4+1} = (-1)^{5} \frac{4}{5} = -1 \times \frac{4}{5} = -\frac{4}{5}\).
06

Calculating the 100th Term

For the 100th term \(a_{100}\), substitute \(n = 100\) into the formula. Thus, \(a_{100} = (-1)^{100+1} \frac{100}{100+1} = (-1)^{101} \frac{100}{101} = -1 \times \frac{100}{101} = -\frac{100}{101}\).

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Key Concepts

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

Understanding the Sequence Formula
An algebraic sequence can be defined by a specific formula that allows us to calculate any term in the sequence based purely on the position of the term, denoted as \( n \). Consider the sequence given by the formula: \[a_n = (-1)^{n+1} \frac{n}{n+1}. \]This formula does a couple of things:
  • The expression \((-1)^{n+1}\) gives us a sign change with each subsequent term. This is essential to introduce the element of an alternating sequence.
  • The fraction \(\frac{n}{n+1}\) determines the actual value of each term, increasing with each increase in \( n \).
By substituting different values for \( n \), you can generate as many terms as needed in the sequence. This makes it a powerful tool for exploring and understanding the pattern of the sequence.
Techniques for Calculating Terms
Calculating terms in an algebraic sequence involves substituting the position of the term, denoted as \( n \), into the given formula. Let's break down the process using the sequence formula discussed earlier.
To find any term \( a_n \) in the sequence:
  • Start by substituting the desired position \( n \) into the formula.
  • Simplify the power \((-1)^{n+1}\) which alternates signs from positive to negative through subsequent terms.
  • Calculate the fraction \(\frac{n}{n+1}\) to finalize the term's value.
      • Examples:
      • For \( n = 1 \), \( a_1 = \frac{1}{2} \), the fraction is positive, since \((-1)^2 = 1\).
      • For \( n = 100 \), \( a_{100} = -\frac{100}{101} \), note the sign change, as \((-1)^{101} = -1\).
      This approach allows you to systematically determine any term within the sequence, providing insight into the sequence's behavior.
Patterns in an Alternating Sequence
An alternating sequence is a sequence where the signs of the terms switch in a regular pattern. In our given sequence, this alternation is driven by the expression \((-1)^{n+1}\). Let's delve into how this impacts the sequence:
  • The sequence alternates signs because the power of -1 oscillates between even and odd as \( n \) increments. Odd powers result in negative products, while even powers yield positive products.
  • The repetition is every other term, which means every two terms, the sign flips. This results in a predictable wave-like pattern.
  • The fraction \(\frac{n}{n+1}\) acts more subtly, increasing the magnitude of the terms but adhering to the alternating sign rule.
Understanding these patterns gives us the tools to predict the nature of future terms without even calculating them fully. Such insight is valuable in making conjectures or predictions for larger sets of terms in algebraic sequences.

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Most popular questions from this chapter

Exponentials of an Arithmetic Sequence If \(a_{1}, a_{2},\) \(a_{3}, \ldots\) is an arithmetic sequence with common difference \(d,\) show that the sequence $$10^{a_{1}}, 10^{a_{2}}, 10^{a_{3}}, \ldots$$ is a geometric sequence, and find the common ratio.

Geometry The midpoints of the sides of a square of side 1 are joined to form a new square. This procedure is repeated for each new square. (See the figure.) (a) Find the sum of the areas of all the squares. (b) Find the sum of the perimeters of all the squares.

Mortgage A couple secures a 30 -year loan of \(\$ 100,000\) at 9\(\frac{3}{4} \%\) per year, compounded monthly, to buy a house. (a) What is the amount of their monthly payment? (b) What total amount will they pay over the 30 -year period? (c) If, instead of taking the loan, the couple deposits the monthly payments in an account that pays 9\(\frac{3}{4} \%\) interest per year, compounded monthly, how much will be in the account at the end of the 30 -year period?

Find the 24 th term in the expansion of \((a+b)^{25}\)

Arithmetic or Geometric? The first four terms of a sequence are given. Determine whether these terms can be the terms of an arithmetic sequence, a geometric sequence, or neither. Find the next term if the sequence is arithmetic or geometric. (a) \(5,-3,5,-3, \ldots \quad\) (b) \(\frac{1}{3}, 1, \frac{5}{3}, \frac{7}{3}, \ldots\) (c) \(\sqrt{3}, 3,3 \sqrt{3}, 9, \ldots\) (d) \(1,-1,1,-1, \ldots\) \(\begin{array}{ll}{\text { (e) } 2,-1, \frac{1}{2}, 2, \ldots} & {\text { (f) } x-1, x, x+1, x+2, \ldots} \\ {\text { (g) }-3,-\frac{3}{2}, 0, \frac{3}{2}, \ldots} & {\text { (h) } \sqrt{5}, \sqrt[3]{5}, \sqrt[6]{5}, 1, \ldots}\end{array}\)
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