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Chapter 8: Problem 76

Express each sequence \(\left\\{a_{n}\right\\}_{n=1}^{\infty}\) as an equivalent sequence of the form \(\left\\{b_{n}\right\\}_{n=3}^{\infty}\). $$\\{2 n+1\\}_{n=1}^{\infty}$$

Short Answer

Expert verified
Answer: The equivalent sequence is \(\left\{2n+5\right\}_{n=3}^{\infty}\).

Step by step solution

01

Analyze the given sequence

The given sequence is \(\left\{2n+1\right\}_{n=1}^{\infty}\). This is an arithmetic sequence with the first term, \(n = 1\), as \(2(1)+1=3\), and the common difference as \(2\). The sequence progresses as \(3, 5, 7, 9, \ldots\).
02

Express the original sequence starting from index 3

We need to find the relation between the original sequence and the new sequence that starts from index 3. Essentially, we want to find the sequence \(\left\{b_n\right\}_{n=3}^{\infty}\) that corresponds to the original sequence values starting from \(n = 3\). In other words, we want \(b_3 = a_3, b_4 = a_4, b_5 = a_5, \ldots\). To find the expression for the new sequence, let's try to find a relation between \(n\) and \(b_n\). We can begin by comparing the indices of the two sequences. The original sequence starts at index \(1\), while the new sequence starts at index \(3\). Therefore, the new sequence index is \(2\) units ahead of the original sequence index: \(n_{new} = n_{old} + 2\). Now, let's substitute this relation into the original sequence expression: \(b_n = 2(n + 2) + 1\).
03

Simplify the expression for the new sequence

Simplify the expression \(b_n = 2(n + 2) + 1\) as follows: \begin{align*} b_n &= 2(n + 2) + 1 \\ &= 2n + 4 + 1 \\ &= 2n + 5 \end{align*}
04

Write the final equivalent sequence

Now we have the equivalent sequence in the required form, so we can write the final answer as: $$\left\{b_n\right\}_{n=3}^{\infty} = \left\{2n+5\right\}_{n=3}^{\infty}$$ Thus, the sequence \(\left\{2n+1\right\}_{n=1}^{\infty}\) is equivalent to the sequence \(\left\{2n+5\right\}_{n=3}^{\infty}\).

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Key Concepts

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

Arithmetic Sequence
An arithmetic sequence is a list of numbers where the difference between successive terms is constant. This fixed number is known as the 'common difference' of the sequence. The simplest arithmetic sequence is formed by adding the common difference to the first term repeatedly to get the rest of the terms. For instance, in the sequence \(3, 5, 7, 9, \ldots\), the first term is 3, and the common difference is 2. Subsequent terms are found by adding 2 to the previous term.

To write a formula for the nth term of an arithmetic sequence, we use \(a_n = a_1 + (n - 1)d\), where \(a_n\) represents the nth term, \(a_1\) is the first term, \(n\) is the term number, and \(d\) is the common difference. We can use this formula to find any term in the sequence, showing that arithmetic sequences can be characterized by linear expressions.
Sequence Transformation
Sequence transformation is the process of changing one sequence into another, while maintaining a certain equivalence between them. The goal is to articulate a relationship between the terms of the original sequence and those of the transformed sequence.

When transforming an arithmetic sequence, a common strategy is to adjust the formula to start at a different index. For example, if we have an arithmetic sequence starting with index 1, and we wish to express it starting from index 3, we need to find the relation between the indices, as seen in our problem. The transformed sequence should maintain the same arithmetic properties as the initial sequence, hence the equivalence of the two sequences in terms of their elements. In our case, the original sequence started with \(n=1\), and we were asked to find the transformed sequence starting with \(n=3\). By adjusting the index and applying the relation to the original formula, we effectively transformed the sequence.
Mathematical Induction
Mathematical induction is a proof technique used to demonstrate the truth of a statement for all natural numbers. It comprises two steps: the base case and the induction step. The base case verifies that the statement holds for the first natural number, usually \(n=1\). The induction step proves that if the statement holds for some natural number \(n\), it also holds for \(n+1\).

For sequences, induction can be used to prove properties or formulas related to the terms of the sequence. For example, one could use induction to show that a given sequence formula correctly generates its terms. In the current context, we are not directly using mathematical induction, but understanding this method provides a solid foundation for grasping the proofs and principles underlying sequence properties.

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Most popular questions from this chapter

The sequence \(\\{n !\\}\) ultimately grows faster than the sequence \(\left\\{b^{n}\right\\},\) for any \(b>1,\) as \(n \rightarrow \infty .\) However, \(b^{n}\) is generally greater than \(n !\) for small values of \(n\). Use a calculator to determine the smallest value of \(n\) such that \(n !>b^{n}\) for each of the cases \(b=2, b=e,\) and \(b=10\).

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{3 n^{2}}{4 n^{2}+1}=\frac{3}{4}$$

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. Jack took a \(200-\mathrm{mg}\) dose of a painkiller at midnight. Every hour, \(5 \%\) of the drug is washed out of his bloodstream. Let \(d_{n}\) be the amount of drug in Jack's blood \(n\) hours after the drug was taken, where \(d_{0}=200 \mathrm{mg}\)

A tank is filled with 100 L of a \(40 \%\) alcohol solution (by volume). You repeatedly perform the following operation: Remove 2 L of the solution from the tank and replace them with 2 L of \(10 \%\) alcohol solution. a. Let \(C_{n}\) be the concentration of the solution in the tank after the \(n\) th replacement, where \(C_{0}=40 \% .\) Write the first five terms of the sequence \(\left\\{C_{n}\right\\}\). b. After how many replacements does the alcohol concentration reach \(15 \% ?\). c. Determine the limiting (steady-state) concentration of the solution that is approached after many replacements.

Consider the geometric series \(S=\sum_{k=0}^{\infty} r^{k}\) which has the value \(1 /(1-r)\) provided \(|r|<1\). Let \(S_{n}=\sum_{k=0}^{n-1} r^{k}=\frac{1-r^{n}}{1-r}\) be the sum of the first \(n\) terms. The magnitude of the remainder \(R_{n}\) is the error in approximating \(S\) by \(S_{n} .\) Show that $$ R_{n}=S-S_{n}=\frac{r^{n}}{1-r} $$
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