/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Find an expression for \(a_{n}\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find an expression for \(a_{n}\) on the basis of the values of \(a_{0}, a_{1}, a_{2}, \ldots\) \(2,0,2,0,2\)

Short Answer

Expert verified
The expression for \(a_n\) is \(a_n = 2(1 - (n \mod 2))\).

Step by step solution

01

Identify the Pattern

Observe the given sequence: \(2, 0, 2, 0, 2, \ldots\). Notice that the sequence is repeating every two terms: it alternates between 2 and 0.
02

Define the Sequence Terms

Define the terms based on their position in the sequence. When the position \(n\) is even, the term is 2. When \(n\) is odd, the term is 0.
03

Express the Terms Mathematically

We use the remainder when division is applied (also known as the modulus operation) to express formulas for each case. When \(n\) is even, the condition is \(n \mod 2 = 0\), yields \(a_n = 2\). When \(n\) is odd, since \(n \mod 2 = 1\), the condition yields \(a_n = 0\).
04

Combine into a Single Formula

Combine these observations into a single formula using the modulus operation: \[a_n = 2(1 - (n \mod 2))\]This works because when \(n\) is even, \((n \mod 2) = 0\) so \(a_n = 2(1 - 0) = 2\), and when \(n\) is odd, \((n \mod 2) = 1\) so \(a_n = 2(1 - 1) = 0\).

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Key Concepts

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

Patterns in Sequences
In mathematics, a sequence is essentially an ordered list of numbers, and finding patterns within them can help us predict subsequent numbers in the sequence. By observing patterns, we can often derive a general formula for the sequence, which can be used to find any term. Take the sequence in the exercise: 2, 0, 2, 0, 2. You might notice immediately that it alternates between 2 and 0. Identifying this pattern is the first step in understanding the sequence.

Recognizing repetitive patterns is crucial. For instance, like in this sequence, patterns that repeat every two numbers are common. Frequently, these patterns can be described using arithmetic operations or mathematical functions. Understanding the structure of a sequence allows us to write its general formula, making predictions about future terms straightforward.
Modulus Operation
The modulus operation is a fascinating mathematical concept often used in sequence problems. It gives the remainder of a division of one number by another. For example, when we divide 5 by 2, the quotient is 2, and the remainder is 1, so 5 mod 2 equals 1. In sequence patterns, the modulus is especially useful in determining even or odd positions.

In the case of the sequence 2, 0, 2, 0, 2, the modulus helps dictate whether a term is even or odd. When applying it to the sequence formula, \(n \mod 2\) distinguishes between even and odd positions within the sequence. For example:
  • If \(n \mod 2 = 0\), \(n\) is even.
  • If \(n \mod 2 = 1\), \(n\) is odd.

This use of modulus allows us to switch the sequence terms efficiently, helping us write succinct formulas like \[a_n = 2(1 - (n \mod 2))\].
Even and Odd Numbers
Being able to distinguish between even and odd numbers is fundamental in mathematics. An even number is any integer divisible by 2 with no remainder, like 0, 2, 4, etc. Conversely, odd numbers have a remainder of 1 when divided by 2, such as 1, 3, 5, etc.

In sequences, identifying whether a number is even or odd can determine which mathematical rule or pattern applies. For example, in alternating sequences (like 2, 0, 2, 0, 2), even positions (0, 2, 4, etc.) might have different formulas than odd positions (1, 3, 5, etc.).

Mastering the ability to see which position is even and which is odd allows for efficient application of formulas and ensures accurate predictions when working with sequences and series, whether you're doing homework or tackling more complex problems.

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Most popular questions from this chapter

Because of complex interactions with other drugs, some drugs have zeroth order elimination kinetics in some circumstances, and first order kinetics in other circumstances, depending on what other drugs are in the patient's system, as well as on age and preexisting medical conditions. Use the data on how concentration varies with time to determine whether the drug has zeroth or first order kinetics. Given the following sequence of measurements for drug concentration, determine whether the drug has zeroth or first order kinetics. $$ \begin{array}{lcccc} \hline \boldsymbol{t} \text { (Hours) } & 0 & 1 & 2 & 3 \\ \hline \boldsymbol{c}_{t} \text { (mg/liter) } & 16 & 12 & 9 & 6.75 \\ \hline \end{array} $$

A population obeys the Beverton-Holt model. You know that \(R_{0}=3\) for this population. As \(t \rightarrow \infty\) you observe that \(N_{t} \rightarrow 100 .\) What value of \(a\) is needed in the model to fit it to these data?

Adderall is a proprietary combination of amphetamine salts that is used to treat ADHD (AttentionDeficit/Hyperactivity Disorder). The patient takes one pill before \(8 \mathrm{am}\), and the drug has completely entered the blood by \(8 \mathrm{am} .\) At 8 am the blood concentration of the drug is \(33.8 \mathrm{ng} / \mathrm{ml}\). Adderall has first order elimination kinetics with \(7.7 \%\) of the drug being removed from the blood in each hour. The data in this equation are taken from Greenhill et al. (2003). (a) The concentration of drug in the patient's blood \(t\) hours after 8 am is measured to be \(C_{t}\). Write a recursive relation for \(C_{t+1}\) in terms of \(C_{t}\). Assume that the patient takes no other Adderall pills. (b) Solve your recurrence equation to derive an explicit formula for \(C_{t}\) as a function of \(t\). (c) When does the concentration of drug first drop below \(0.1 \mathrm{ng} / \mathrm{ml} ?\)

Use the formal definition of limits to show that \(\lim _{n \rightarrow \infty} a_{n}=a ;\) that is, find \(N\) such that for every \(\epsilon>0\), there exists an \(N\) such that \(\left|a_{n}-a\right|<\epsilon\) whenever \(n>N\). $$ \lim _{n \rightarrow \infty} e^{-2 n}=0 $$

Use the formal definition of limits to show that \(\lim _{n \rightarrow \infty} a_{n}=a ;\) that is, find \(N\) such that for every \(\epsilon>0\), there exists an \(N\) such that \(\left|a_{n}-a\right|<\epsilon\) whenever \(n>N\). $$ \lim _{n \rightarrow \infty} \frac{1}{n+1}=0 $$

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