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The given pattern continues. Write down the nth term of a sequence \(\left\\{a_{n}\right\\}\) suggested by the pattern. \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots\)

Short Answer

Expert verified
The nth term is \(a_{n} = \frac{n}{n+1}\).

Step by step solution

01

Identify the Numerator Pattern

Observe the numerical pattern of the sequence: \ \ \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots \). The numerators are 1, 2, 3, 4, ... and so on. Thus, the numerator for the nth term is simply n.
02

Identify the Denominator Pattern

Next, observe the denominators of the sequence: \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots \). The denominators are 2, 3, 4, 5, ... which are n + 1 for each corresponding numerator. Therefore, the denominator for the nth term is n + 1.
03

Combine the Patterns

Combine the patterns identified for the numerator and the denominator. The nth term of the sequence can be expressed as: \ \ \(a_{n} = \frac{n}{n+1}\)
04

Write the General Term

Therefore, the general term for the sequence is: \ \ \(a_{n} = \frac{n}{n+1}\)

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Key Concepts

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

Numerator Pattern
Identifying the numerator pattern in a sequence is crucial for understanding how to find the nth term. When looking at the sequence \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots\), observing the numerators is the first step.

The numerators in this sequence are clearly following a simple arithmetic pattern: 1, 2, 3, 4,\ldots. Notice how each term increases by 1 compared to the previous term. This consistent increase by 1 implies that the numerator for the nth term is simply n.

So, for our sequence, the numerator of the nth term can be denoted as \(n\).
Denominator Pattern
Next, we need to identify the pattern in the denominators, which can sometimes be a bit trickier. Examine the given sequence again: \(\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots\).

We see another clear pattern: the denominators are 2, 3, 4, 5, \ldots. To find a relation to the numerators, compare each numerator and its corresponding denominator:
  • The denominator of the first term is 2 (1 + 1).
  • The denominator of the second term is 3 (2 + 1).
  • The denominator of the third term is 4 (3 + 1).
The denominators are always one more than their respective numerators. Thus, the denominator for the nth term in the sequence can be expressed as \(n + 1\).
General Term Formula
After identifying the patterns in both the numerators and the denominators, we can combine these to find the general term formula.

From our observations:
  • The numerator of the nth term is \(n\).
  • The denominator of the nth term is \(n + 1\).
Therefore, the nth term of the sequence can be written as \(a_{n} = \frac{n}{n+1}\).

This formula is crucial as it provides a simple and concise method to find any term in the sequence, without having to list all the preceding terms. To reinforce understanding, you can plug in different values for n and watch the pattern hold:
  • For \(n = 1\), \(a_{1} = \frac{1}{2}\).
  • For \(n = 2\), \(a_{2} = \frac{2}{3}\).
  • For \(n = 3\), \(a_{3} = \frac{3}{4}\).
Understanding these patterns and how they combine into the general term formula allows a deeper grasp of sequence behavior.

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Most popular questions from this chapter

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Bode's Law In \(1772,\) Johann Bode published the following formula for predicting the mean distances, in astronomical units (AU), of the planets from the sun: $$ a_{1}=0.4 \quad a_{n}=0.4+0.3 \cdot 2^{n-2} $$ where \(n \geq 2\) is the number of the planet from the sun. (a) Determine the first eight terms of the sequence. (b) At the time of Bode's publication, the known planets were Mercury \((0.39 \mathrm{AU}),\) Venus \((0.72 \mathrm{AU}),\) Earth \((1 \mathrm{AU})\) Mars \((1.52 \mathrm{AU}),\) Jupiter \((5.20 \mathrm{AU}),\) and Saturn \((9.54 \mathrm{AU})\) How do the actual distances compare to the terms of the sequence? (c) The planet Uranus was discovered in \(1781,\) and the asteroid Ceres was discovered in \(1801 .\) The mean orbital distances from the sun to Uranus and Ceres " are \(19.2 \mathrm{AU}\) and \(2.77 \mathrm{AU},\) respectively. How well do these values fit within the sequence? (d) Determine the ninth and tenth terms of Bode's sequence. (e) The planets Neptune and Pluto" were discovered in 1846 and \(1930,\) respectively. Their mean orbital distances from the sun are \(30.07 \mathrm{AU}\) and \(39.44 \mathrm{AU},\) respectively. How do these actual distances compare to the terms of the sequence? (f) On July \(29,2005,\) NASA announced the discovery of a dwarf planet \((n=11),\) which has been named Eris. Use Bode's Law to predict the mean orbital distance of Eris from the sun. Its actual mean distance is not yet known, but Eris is currently about 97 astronomical units from the sun.

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