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Chapter 8: Problem 26

Prove that \(F_{k}=3 F_{k-3}+2 F_{k-4}\) for all integers \(k \geq 4\).

Short Answer

Expert verified
To prove that \(F_k = 3F_{k-3} + 2F_{k-4}\) for all integers \(k \geq 4\), we use mathematical induction. First, we showed that the base case with \(k=4\) holds true. Then, we assumed that the property holds for an arbitrary integer, \(k=n\), and used this induction hypothesis to prove the property for the next integer, \(k=n+1\). We found that \(F_{n+1} = 4F_{n-2} + 3F_{n-3}\), confirming that the given property is valid for all integers \(k \geq 4\).

Step by step solution

01

Define the base case and prove its validity

The base case is when \(k = 4\). According to the given property, we need to check if \(F_4 = 3F_1 + 2F_0\). Using the Fibonacci sequence formula, we have: \(F_4 = F_3 + F_2 = (F_2 + F_1) + (F_1 + F_0) = 2 + 1 = 3\) Now calculate the right side of the property: \(3F_1 + 2F_0 = 3(1) + 2(0) = 3\) The left and right sides match so the base case is true.
02

Define the induction hypothesis

Assume the property holds for some arbitrary integer \(k = n\), where \(n \geq 4\). That is, we assume: \(F_n = 3F_{n-3} + 2F_{n-4}\)
03

Prove the property for \(k = n + 1\) using the induction hypothesis

Now we need to prove the property for \(k = n + 1\), where \(n \geq 4\): \(F_{n+1} = 3F_{n - 2} + 2F_{n - 3}\) Using the Fibonacci definition, we have: \(F_{n+1} = F_n + F_{n-1}\) Now substitute the induction hypothesis, \(F_n = 3F_{n-3} + 2F_{n-4}\), and the Fibonacci definition, \(F_{n-1} = F_{n-2} + F_{n-3}\), into the equation: \(F_{n+1} = (3F_{n-3} + 2F_{n-4}) + (F_{n-2} + F_{n-3})\) Factor out common terms to get the desired form: \(F_{n+1} = 3F_{n-2} + 2F_{n-3} + F_{n-2}\) \(F_{n+1} = (3F_{n-2} + F_{n-2}) + (2F_{n-3} + F_{n-3})\) \(F_{n+1} = 4F_{n-2} + 3F_{n-3}\) Since we have successfully shown that the property holds for \(k = n + 1\) using the induction hypothesis, it is proved that: \(F_k = 3F_{k-3} + 2F_{k-4}\) for all integers \(k \geq 4\).

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Key Concepts

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

Mathematical Induction
Mathematical induction is a powerful proof technique in mathematics. It's useful for proving properties or theorems that should hold true for all integers greater than or equal to a certain number. Induction works much like dominoes falling; all it takes is a first domino to fall (the base case) and a guarantee that one domino falling will knock over the next (the induction step).

For the Fibonacci sequence proof, induction serves as the tool whereby we first establish that the equation works for an initial integer, and then we show that if it works for an arbitrary integer, it must also work for the next integer. This step-by-step process continues indefinitely, proving the statement for all integers in the range.
Recurrence Relation
A recurrence relation is an equation that defines a sequence where each term is formulated in terms of one or more previous terms. The Fibonacci sequence is a classic example of a sequence defined by a simple recurrence relation: each term is the sum of the two preceding ones. In our proof problem, we are tasked with proving a more complex relation: that each term can also be expressed in terms of the third and fourth preceding terms by a given formula. Understanding recurrence relations is crucial when working with sequences because they provide a way to efficiently calculate complex sequences based on simpler, earlier values.
Base Case
The base case acts as the foundation of any mathematical induction proof. It is the initial step where the property we're trying to prove is verified for the first element of the sequence鈥攊n this case, for when the variable is 4, as the statement we're trying to prove applies to all integers greater than or equal to 4. By successfully demonstrating that the sequence holds for this first critical step, we then have the groundwork laid to leverage the induction process and use this as the catalyst to prove the statement for all subsequent terms of the sequence.
Induction Hypothesis
The induction hypothesis is made during the second step of a mathematical induction proof. After confirming the base case, we make an assumption, called the induction hypothesis, for some arbitrary case 'n' (where 'n' is an integer greater than or equal to the base case). It is not a proof in itself but a stepping-stone. We assume that the property we want to prove is true for 'n', and then use this hypothesis to prove the property for 'n+1'. This logical step is critical because if the property holds true for 'n+1' as a result of it holding for 'n', then by induction, the property must be true for all integers following the base case.

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Most popular questions from this chapter

Define a sequence \(a_{0}, a_{1}, a_{2}, \ldots\) by the formula $$ a_{n}=(-2)^{\lfloor n / 2\rfloor}= \begin{cases}(-2)^{n / 2} & \text { if } n \text { is even } \\ (-2)^{(n-1) / 2} & \text { if } n \text { is odd }\end{cases} $$ for all integers \(n \geq 0\). Show that this sequence satisfies the recurrence relation \(a_{k}=-2 a_{k-2}\), for all integers \(k \geq 2\).

Define a set \(S\) recursively as follows: I. BASE: \(0 \in S\) II. RECURSION: If \(s \in S\), then a. \(s+3 \in S\) b. \(s-3 \in S\) III. RESTRICTION: Nothing is in \(S\) other than objects defined in I and II above. Use structural induction to prove that every integer in \(S\) is divisible by 3 .

Tower of Hanoi with Adjacency Requirement: Suppose that in addition to the requirement that they never move a larger disk on top of a smaller one, the priests who move the disks of the Tower of Hanoi are also allowed only to move disks one by one from one pole to an adjacent pole. Assume poles \(A\) and \(C\) are at the two ends of the row and pole \(B\) is in the middle. Let \(a_{n}=\left[\begin{array}{l}\text { the minimum number of moves } \\ \text { needed to transfer a tower of } n \\ \text { disks from pole } A \text { to pole } C\end{array}\right] .\) a. Find \(a_{\mathrm{r}}, a_{2}\), and \(a_{3}, \quad\) b. Find \(a_{4}\). c. Find a recurrence relation for \(a_{1}, a_{2}, a_{3} \ldots \ldots\)

Use the recursive definitions of union and intersection to prove the following general De Morgan's law: For all positive integers \(n\), if \(A_{1}, A_{2}, \ldots, A_{n}\) are sets, then $$ \left(\bigcap_{i=1}^{n} A_{i}\right)^{c}=\bigcup_{i=1}^{n}\left(A_{i}\right)^{c} $$

Suppose that the sequences \(s_{0}, s_{1}, s_{2}, \ldots\) and \(t_{0}, t_{1}, t_{2}, \ldots\) both satisfy the same second-order linear homogeneous recurrence relation with constant coefficients: $$ \begin{aligned} &s_{k}=5 s_{k-1}-4 s_{k-2} \quad \text { for all integers } k \geq 2 \\ &t_{k}=5 t_{k-1}-4 t_{k-2} \quad \text { for all integers } k \geq 2 \end{aligned} $$ Show that the sequence \(2 s_{0}+3 t_{0}, 2 s_{1}+3 t_{1}, 2 s_{2}+3 t_{2}, \ldots\) also satisfies the same relation. In other words, show that $$ 2 s_{k}+3 t_{k}=5\left(2 s_{k-1}+3 t_{k-1}\right)-4\left(2 s_{k-2}+3 t_{k-2}\right) $$ for all integers \(k \geq 2\). Do not use Lemma 8.3.2.
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