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Chapter 9: Problem 41

Write an expression for the apparent \(n\) th term \(\left(a_{n}\right)\) of the sequence. (Assume that \(n\) begins with 1.) $$\frac{2}{1}, \frac{3}{3}, \frac{4}{5}, \frac{5}{7}, \frac{6}{9}, \dots$$

Short Answer

Expert verified
The nth term of the sequence is \( \frac{n+1}{2n-1} \).

Step by step solution

01

Identifying the Numerator Pattern

The pattern in the numerator is that each term equals its index: the first term is 2, the second term is 3, and so forth. Thus, the numerator of the nth term should be \( n+1 \).
02

Identifying the Denominator Pattern

The pattern in the denominator is that each term equals 2n-1: the first term is 1, the second term is 3, the third term is 5, and so forth, each time increasing by 2. Thus, the denominator of the nth term should be \( 2n-1 \).
03

Formulating the nth Term

Now, combine the two patterns to formulate the nth term of the sequence. The numerator is \( n+1 \) and the denominator is \( 2n-1 \). Therefore, the nth term, \( a_n \), of the sequence is \( \frac{n+1}{2n-1} \).

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Key Concepts

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

nth term
In sequences, the "nth term" is a key concept that allows us to determine any specific term in the sequence based on its position. Given a sequence, the nth term formula helps in predicting the value of the sequence at step "n" without having to calculate all previous terms. This is particularly useful for long sequences.
To find the nth term of a sequence like \( \frac{2}{1}, \frac{3}{3}, \frac{4}{5}, \frac{5}{7}, \frac{6}{9}, \dots \), we need to first understand the underlying patterns of both the numerators and denominators. Once these patterns are identified, they are expressed in terms of "n", to formulate the nth term of the sequence, \( a_n \), using the patterns observed.
patterns in sequences
Finding patterns in sequences is essential to solving problems related to them. Patterns help us understand what changes between each term and help in predicting future elements of the sequence.
In the provided sequence, each numerator increases by one with each step. This means the numerator pattern can be described as "increasing by one" starting from 2, which is captured by the expression \( n+1 \).
Similarly, each denominator in the sequence increases by 2 upon moving to the next term. Starting with 1, the pattern can be captured by the rule \( 2n-1 \).
Identifying these patterns is crucial as it simplifies the process of formulating the nth term, which reveals the structure and behavior of the entire sequence.
numerator and denominator relationships
Understanding the relationship between numerators and denominators in a sequence is vital, as they often display distinct yet interconnected patterns.
In our given sequence \( \frac{2}{1}, \frac{3}{3}, \frac{4}{5}, \frac{5}{7}, \dots \), the relationship between numerators and denominators is highlighted by their individual growth patterns. The numerator, \( n+1 \), follows a linear progression as it increases by one with each new sequence term. Meanwhile, the denominator, \( 2n-1 \), grows at a different pace, increasing by two each time.
By acknowledging these patterns, we uncover the perfect balance needed to construct the nth term \( a_n = \frac{n+1}{2n-1} \) which reflects the distinct yet harmonious progression of both elements in the sequence.

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Most popular questions from this chapter

Consider \(n\) independent trials of an experiment in which each trial has two possible outcomes: "success" or "failure." The probability of a success on each trial is \(p,\) and the probability of a failure is \(q=1-p .\) In this context, the term \(_{n} C_{k} p^{k} q^{n-k}\) in the expansion of \((p+q)^{n}\) gives the probability of \(k\) successes in the \(n\) trials of the experiment.To find the probability that the sales representative in Exercise 87 makes four sales when the probability of a sale with any one customer is \(\frac{1}{2},\) evaluate the term $$_{8} C_{4}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{4}$$, in the expansion of \(\left(\frac{1}{2}+\frac{1}{2}\right)^{8}\).

Form rows \(8-10\) of Pascal's Triangle.

Use the Binomial Theorem to expand the complex number. Simplify your result. $$(1+i)^{4}$$

Prove the identity. $$_{n} C_{r}=\frac{_{n} P_{r}}{r !}$$

The table shows the average prices \(f(t)\) (in cents per kilowatt hour) of residential electricity in the United States from 2003 through 2010 . (Source: U.S. Energy Information Administration ).$$\begin{array}{|c|c|}\hline \text { Year } & \text { Abcrage Poids }(10) \\\\\hline 2003 & 8.72 \\\2004 & 8.95 \\\2005 & 9.45 \\\2006 & 10.40 \\\2007 & 10.65 \\\2008 & 11.26 \\\2009 & 11.51 \\\2010 & 11.58 \\\\\hline\end{array}$$.(a) Use the regression feature of a graphing utility to find a cubic model for the data. Let \(t\) represent the year, with \(t=3\) corresponding to 2003 (b) Use the graphing utility to plot the data and the model in the same viewing window. (c) You want to adjust the model so that \(t=3\) corresponds to 2008 rather than \(2003 .\) To do this, you shift the graph of \(f\) five units to the left to obtain \(g(t)=f(t+5) .\) Use binomial coefficients to write \(g(t)\) in standard form. (d) Use the graphing utility to graph \(g\) in the same viewing window as \(f\) (e) Use both models to predict the average price in \(2011 .\) Do you obtain the same answer? (f) Do your answers to part (e) seem reasonable? Explain. (g) What factors do you think may have contributed to the change in the average price?
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