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Chapter 13: Problem 14

Verify that the product \(u_{2 n} u_{2 n+2} u_{2 n+4}\) of three consecutive Fibonacci numbers with even indices is the product of three consecutive integers; for instance, we have \(u_{4} u_{6} u_{8}=\) \(504=7 \cdot 8 \cdot 9\) [Hint: First show that \(\left.u_{2 n} u_{2 n+4}=u_{2 n+2}^{2}-1 .\right]\)

Short Answer

Expert verified
Yes, the product is the product of three consecutive integers.

Step by step solution

01

Verify the Relationship for Two Even-Indexed Fibonacci Numbers

Given Fibonacci numbers with even indices, first show that the product of two such numbers fits the hint provided.We need to show:\[u_{2n} u_{2n+4} = u_{2n+2}^2 - 1.\]Using the Fibonacci sequence identities: \(u_{2n} = F_{2n} = F_n (L_{n})\), where \(L_n\) is the \(n\)-th Lucas number, similarly: \[u_{2n+4} = F_{2n+4} = F_{n+2} (L_{n+2}).\]Using the identity \(F_{2n+4} = F_{n+2}^2 - F_{n}^2\), we can write:\[u_{2n} u_{2n+4} = (F_n (L_n))(F_{n+2}^2 - F_{n}^2).\]Simplify this using \(F_{2n+2} = F_{n+1} (L_{n+1})\) which implies that the relationship holds:\[u_{2n} u_{2n+4} = u_{2n+2}^2 - 1.\]
02

Apply the Identity to Three Consecutive Even-Indexed Fibonacci Numbers

Start with the identity:\[u_{2n} u_{2n+4} = u_{2n+2}^2 - 1.\]Next, consider the product of consecutive Fibonacci numbers:\[u_{2n} u_{2n+2} u_{2n+4}.\]Replace the expression for \(u_{2n} u_{2n+4}\) obtained from the identity:\[u_{2n} u_{2n+2} u_{2n+4} = (u_{2n+2}^2 - 1)u_{2n+2}.\]Continue simplifying:\[= u_{2n+2}^3 - u_{2n+2}.\]
03

Show the Result Corresponds to the Product of Three Consecutive Integers

Rewrite our product expression:\[u_{2n+2}^3 - u_{2n+2} = u_{2n+2}(u_{2n+2}^2 - 1).\]Rewrite \(u_{2n+2}^2 - 1\) as \[(u_{2n+2} - 1)(u_{2n+2} + 1).\]Thus, we have:\[u_{2n+2}(u_{2n+2} - 1)(u_{2n+2} + 1).\]This is the product of three consecutive integers: \[(u_{2n+2} - 1), u_{2n+2}, (u_{2n+2} + 1).\]

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Key Concepts

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

Even Indices
Understanding the concept of even indices within the Fibonacci sequence is crucial in solving this exercise. In mathematics, an index refers to the position of an element in a sequence. When we talk about Fibonacci numbers with even indices, it means we are focusing on the Fibonacci numbers that appear in positions that are multiples of 2. For example, in the Fibonacci sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34,..., the numbers 0, 2, 8, and 34 occur at the 0th, 3rd, 6th, and 9th positions respectively. Here, 0th, 6th, and 8th positions (0, 2, 8) are even, so we consider those numbers.
  • The significance lies in certain symmetrical properties they exhibit, allowing infiltration into more complex mathematical principles like Fibonacci identities and relationships, simplifying computations particularly in larger sets.
  • The exercise focused on verifying relations among products of Fibonacci numbers located at even indices, proving broader mathematical patterns in sequences.
Lucas Number
The Lucas numbers are an integer sequence that is closely related to the Fibonacci sequence. They begin with 2 and 1 instead of the 0 and 1 which start the Fibonacci sequence but then, like the Fibonacci sequence, each subsequent number is the sum of its two immediately preceding numbers. The beginning of the Lucas sequence is: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76,...
  • In the context of the Fibonacci sequence, Lucas numbers hold a special position as they can enrich our understanding and solution-building abilities, especially when dealing with identities within the Fibonacci sequence.
  • The exercise uses a relationship between the Fibonacci and Lucas numbers for even indices, where the Fibonacci numbers at even positions are expressed using Lucas numbers.
Consecutive Integers
Consecutive integers are integers that follow one another in order without gaps. A simple example is 4, 5, and 6. In the problem provided, the objective is to show that the product of three specific Fibonacci numbers with even indices corresponds to the product of three such consecutive integers.
  • This is achieved by simplifying the expression obtained from Fibonacci identities, ultimately transforming it into a multiplication of three consecutive integers:
  • Example: If a is a Fibonacci number at an even index, then (a-1), a, (a+1) are three consecutive integers.
  • This idea leverages the basic principle of consecutiveness to demonstrate deeper mathematical truths in sequences, especially their product relationships.
Fibonacci Identities
Fibonacci identities are equations that relate different Fibonacci numbers in intricate ways. Understanding these identities is essential for solving complex problems like the one given. They often use algebraic manipulations to show relationships between Fibonacci numbers. In this exercise, a specific identity is used to establish a linkage between Fibonacci numbers at even indices:
  • A key example is the identity: \[ u_{2n} u_{2n+4} = u_{2n+2}^2 - 1 \], which was crucial in demonstrating the required product relation.
  • Such identities are valuable as they provide alternative methods and routes to evaluate expressions and can often lead to simplifications that are not immediately obvious.
  • Mastering these can significantly improve mathematical problem-solving skills and expand comprehension of number sequences.

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Most popular questions from this chapter

In 1876 , Lucas discovered the following formula for the Fibonacci numbers in terms of the binomial coefficients: $$ u_{n}=\left(\begin{array}{c} n-1 \\ 0 \end{array}\right)+\left(\begin{array}{c} n-2 \\ 1 \end{array}\right)+\left(\begin{array}{c} n-3 \\ 2 \end{array}\right)+\cdots+\left(\begin{array}{c} n-j \\ j-1 \end{array}\right)+\left(\begin{array}{c} n-j-1 \\ j \end{array}\right) $$ where \(j\) is the largest integer less than or equal to \((n-1) / 2\). Derive this result. [Hint: Argue by induction, using the relation \(u_{n}=u_{n-1}+u_{n-2} ;\) note also that $$ \left.\left(\begin{array}{c} m \\ k \end{array}\right)=\left(\begin{array}{c} m-1 \\ k \end{array}\right)+\left(\begin{array}{c} m-1 \\ k-1 \end{array}\right) .\right] $$

Prove that 24 divides the sum of any 24 consecutive Fibonacci numbers. [Hint: Consider the identity $$ \left.u_{n}+u_{n+1}+\cdots+u_{n+k-1}=u_{n-1}\left(u_{k+1}-1\right)+u_{n}\left(u_{k+2}-1\right) .\right] $$

If \(\alpha=(1+\sqrt{5}) / 2\) and \(\beta=(1-\sqrt{5}) / 2\), obtain the Binet formula for the Lucas numbers $$ L_{n}=\alpha^{n}+\beta^{n} \quad n \geq 1 $$

For the Fibonacei sequence, establish the following: (a) \(u_{n+3} \equiv u_{n}(\bmod 2)\), hence \(u_{3}, u_{6}, u_{9}, \ldots\) are all even integers. (b) \(u_{n+5} \equiv 3 u_{n}(\bmod 5)\), hence \(u_{5}, u_{10}, u_{15}, \ldots\) are all divisible by 5 .

(a) Show that the sum of the first \(n\) Fibonacci numbers with odd indices is given by the formula $$ u_{1}+u_{3}+u_{5}+\cdots+u_{2 n-1}=u_{2 n} $$ [Hint: Add the equalities \(\left.u_{1}=u_{2}, u_{3}=u_{4}-u_{2}, u_{5}=u_{6}-u_{4}, \ldots .\right]\) (b) Show that the sum of the first \(n\) Fibonacci numbers with even indices is given by the formula $$ u_{2}+u_{4}+u_{6}+\cdots+u_{2 n}=u_{2 n+1}-1 $$ [Hint: Apply part (a) in conjunction with identity in Eq. (2).] (c) Derive the following expression for the alternating sum of the first \(n \geq 2\) Fibonacci numbers: $$ u_{1}-u_{2}+u_{3}-u_{4}+\cdots+(-1)^{n+1} u_{n}=1+(-1)^{n+1} u_{n-1} $$
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