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Find the first six terms of the sequence with the elements defined as \(F(0)=5, F(1)=\) \(10,\) and \(F(n)=F(n-1)-2 F(n-2)\) for \(n \geq 2\).

Short Answer

Expert verified
The first six terms are 5, 10, 0, -20, -20, 20.

Step by step solution

01

Identify Initial Terms

The problem provides us with the first two terms of the sequence: \( F(0) = 5 \) and \( F(1) = 10 \). These will be our starting points for finding the subsequent terms.
02

Apply Recurrence Formula for Third Term

To find \( F(2) \), we use the recurrence relation: \( F(n) = F(n-1) - 2 F(n-2) \). Substituting \( n = 2 \), we get \( F(2) = F(1) - 2 F(0) = 10 - 2 \times 5 = 0 \).
03

Calculate Fourth Term

Using the recurrence relation and previously calculated terms, find \( F(3) \): \( F(3) = F(2) - 2 F(1) = 0 - 2 \times 10 = -20 \).
04

Determine Fifth Term

Compute \( F(4) \) using the same method: \( F(4) = F(3) - 2 F(2) = -20 - 2 \times 0 = -20 \).
05

Compute Sixth Term

For \( F(5) \), apply the recurrence formula: \( F(5) = F(4) - 2 F(3) = -20 - 2 \times (-20) = -20 + 40 = 20 \).

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Key Concepts

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

sequence calculation
Sequence calculation involves determining a list of numbers in a specific order. Each term in a sequence often relates to its predecessors based on a specific rule. In the given exercise, we have a sequence defined by a recurrence relation. This means every term after the initial ones is calculated based on specific operations involving previous terms.
In our case, the sequence begins with two known values, known as **initial terms**:
  • The first term, \( F(0) = 5 \).
  • The second term, \( F(1) = 10 \).
To find subsequent terms, we apply a formula repeatedly. The formula provided, \( F(n) = F(n-1) - 2 F(n-2) \) for \( n \geq 2 \), determines each upcoming number in the sequence.
This mathematical approach ensures that each number flows logically from those before it. Calculating terms one after the other, following the relation, helps form a clearer understanding of the pattern or progression in the sequence.
discrete mathematics
Discrete mathematics is a branch of mathematics dealing with countable, distinct elements. It includes topics such as sequences, graphs, and logical statements. In this exercise, we specifically deal with sequences, which are a fundamental concept in discrete structures.
A sequence in discrete mathematics can be thought of as a function, mapping each number in its domain (an initial segment of non-negative integers) to a distinct element, like our terms \( F(0), F(1), \) etc. Each element in a sequence is discrete and separate, emphasizing the idea of individual countable units.
This discipline contrasts with continuous mathematics, which deals with connected structures like curves and surfaces. In our exercise, understanding how each term is derived from previous ones exemplifies a core aspect of discrete methods: step-by-step calculation and logical progression.
problem-solving steps
Approaching a mathematical problem usually involves structured problem-solving steps. Having a strategy in place allows for orderly calculation and logical thinking. Here's a typical approach as evidenced in our sequence problem:
  • **Identify Initial Information:** Start by noting down all the given pieces of information, such as initial terms and any relationships or equations provided.
  • **Apply the Given Formula or Expression:** Using the initial terms, apply the given formula to compute new elements. Each term builds based on the defined relation, ensuring consistency and adherence to the problem guidelines.
  • **Check Your Calculations:** As new terms are calculated, double-check each result to ensure accuracy and correct any errors before moving on.
  • **Consider the Pattern:** Look out for any emerging patterns in the sequence. Sometimes, spotting trends can help with predicting future terms or checking current work for plausibility.
This structured approach makes problem solving more efficient and understandable, especially when dealing with complex mathematical sequences.

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Most popular questions from this chapter

A chain-letter scheme is a famous (and usually illegal) get-rich-quick scheme. A person \(X\) receives a letter with, say, five names on it. \(X\) sends 10 to the person whose name is at the top of the list. \(X\) then deletes that name from the top of the list, adds his or her own name to the bottom of the list, and sends the letter to five "friends," all within one day. In around two weeks, \(X\) is supposed to receive 31,250. Suppose every person who receives the letter follows the instructions (including sending 10 to the person listed first!). Show that if there are only finitely many people, the scheme cannot work (in some sense of "cannot work" that you should make precise). Show that if there are countably infinitely many people, the scheme can work.

Let \(X=\\{a, b\\}\) (a) There are nine partial functions \(F: X \rightarrow X\). List them. (b) There are four functions \(F: X \rightarrow X\). List them. (c) List all \(l-l\) functions \(F: X \rightarrow X\). (d) List all onto functions \(F: X \rightarrow X\).

Prove that: (a) \(0.999999 \ldots 99 \ldots=1\) (b) \(0.34627 \overline{0}=0.34626 \overline{9}\).

Let \(A, B,\) and \(C\) be sets, and let \(F: A \rightarrow C\) and \(G: B \rightarrow C\) be functions. (a) What condition must \(F\) and \(G\) satisfy for \(F \cup G\) to be a function from \(A \cup B\) to \(C ?\) (b) Give conditions on \(A\) and \(B\) such that \(F \cup G\) is a function for every \(F: A \rightarrow C\) and \(G: B \rightarrow C\)

Prove that the function \(F: \mathbb{Z} \rightarrow \mathbb{Z}\) defined as \(F(n)=n+6\) is a bijection.

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