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Chapter 9: Problem 6

Each exercise 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 \(a_{4}\). $$a_{n}=\frac{2^{n}-1}{2^{n}}$$

Short Answer

Expert verified
\( a_1 = \frac{1}{2}, a_2 = \frac{3}{4}, a_3 = \frac{7}{8}, a_4 = \frac{15}{16} \).

Step by step solution

01

Calculate \( a_1 \)

To find \( a_1 \), substitute \( n = 1 \) into the formula. \[ a_1 = \frac{2^1 - 1}{2^1} = \frac{2 - 1}{2} = \frac{1}{2} \] So, \( a_1 = \frac{1}{2} \).
02

Calculate \( a_2 \)

For \( a_2 \), substitute \( n = 2 \) into the formula. \[ a_2 = \frac{2^2 - 1}{2^2} = \frac{4 - 1}{4} = \frac{3}{4} \] So, \( a_2 = \frac{3}{4} \).
03

Calculate \( a_3 \)

To find \( a_3 \), substitute \( n = 3 \) into the formula. \[ a_3 = \frac{2^3 - 1}{2^3} = \frac{8 - 1}{8} = \frac{7}{8} \] Thus, \( a_3 = \frac{7}{8} \).
04

Calculate \( a_4 \)

Finally, find \( a_4 \) by substituting \( n = 4 \) into the formula. \[ a_4 = \frac{2^4 - 1}{2^4} = \frac{16 - 1}{16} = \frac{15}{16} \] So, \( a_4 = \frac{15}{16} \).

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Key Concepts

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

Term Calculation
Term calculation involves determining specific elements in a sequence by substituting given values into the provided formula. This enables us to better understand the behavior and pattern of the sequence as a whole.
Let's take a look at how it works using the given problem:
  • Start by inserting the required value of the position into the sequence formula. Here, it is denoted by \( n \).
  • In the exercise, we calculated terms \( a_1 \), \( a_2 \), \( a_3 \), and \( a_4 \) by substituting \( n = 1 \), \( n = 2 \), \( n = 3 \), and \( n = 4 \), respectively.
  • This process helps us find the first few terms to get an understanding of the development of the sequence.
By following these steps, you can successfully compute any term within a sequence and gain insights into its structure.
Nth Term Formula
The \( n \)th term formula is key to understanding sequences. It provides a rule or equation that allows you to find any term in the sequence without needing to calculate all prior terms.
In this exercise, the \( n \)th term formula given is \( a_n = \frac{2^n - 1}{2^n} \). Let's dive deeper:
  • It involves powers of two, indicating exponential growth.
  • Each \( n \) is substituted into the formula to determine its respective term. By using \( n = 1 \), \( n = 2 \), etc., the specific terms we calculated earlier are obtained.
  • Such a formula gives a concise way to find not just the first term but any term \( a_n \), saving time and effort.
Understanding and interpreting the formula allows us to explore and predict how the sequence behaves beyond just the first few terms.
Mathematical Sequences
A mathematical sequence is an ordered list of numbers following a specific pattern. Recognizing the rules that form sequences is crucial to solving related problems.
For our given sequence, the common pattern is defined by the formula \( a_n = \frac{2^n - 1}{2^n} \). Here are some features of sequences highlighted by this exercise:
  • They often have a starting point, such as \( n = 1 \), though sequences can start from other integers.
  • The terms progress in a logical, predictable manner when following the formula provided.
  • Sequences can demonstrate a variety of behaviors such as growth, decay, oscillation, or stabilization depending on the rule.
Mathematical sequences are everywhere, from simple arithmetic sequences to complex functions in calculus, making them a foundational topic in mathematics.

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Most popular questions from this chapter

a. Use Taylor's formula with \(n=2\) to find the quadratic approximation of \(f(x)=(1+x)^{k}\) at \(x=0\) ( \(k\) a constant). b. If \(k=3,\) for approximately what values of \(x\) in the interval [0,1] will the error in the quadratic approximation be less than \(1 / 100 ?\)

Show that \(\Sigma_{n=1}^{\infty} 2^{(n-1)} / n !\) diverges. Recall from the Laws of Exponents that \(2^{(n)}=\left(2^{n}\right)^{n}\)

Find the values of \(x\) for which the given geometric series converges. Also, find the sum of the series (as a function of \(x\) ) for those values of \(x .\) $$\sum_{n=0}^{\infty}(-1)^{n}(x+1)^{n}$$

Uniqueness of limits Prove that limits of sequences are unique. That is, show that if \(L_{1}\) and \(L_{2}\) are numbers such that \(a_{n} \rightarrow L_{1}\) and \(a_{n} \rightarrow L_{2},\) then \(L_{1}=L_{2}\).

Assume that each sequence converges and find its limit. $$a_{1}=0, \quad a_{n+1}=\sqrt{8+2 a_{n}}$$
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