/*! 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 56 The Fibonacci sequence was defin... [FREE SOLUTION] | 91Ó°ÊÓ

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The Fibonacci sequence was defined in Section 8.1 by the equations $$f_{1}=1, \quad f_{2}=1, \quad f_{n}=f_{n-1}+f_{n-2} \quad n \geqslant 3$$ Show that each of the following statements is true. (a) $$\frac{1}{f_{n-1} f_{n+1}}=\frac{1}{f_{n-1} f_{n}}-\frac{1}{f_{n} f_{n+1}} \quad$$ (b) $$\sum_{n=2}^{x} \frac{1}{f_{n-1} f_{n+1}}=1$$ (c) $$\sum_{n=2}^{\infty} \frac{f_{n}}{f_{n-1} f_{n+1}}=2$$

Short Answer

Expert verified
Parts (a) and (b) are shown using telescoping series, and (c) simplifies to 2.

Step by step solution

01

Understanding Equation (a)

We need to show \( \frac{1}{f_{n-1} f_{n+1}} = \frac{1}{f_{n-1} f_{n}} - \frac{1}{f_{n} f_{n+1}} \). Start by understanding that this is a telescoping fraction identity. It can be rewritten as: \[ \frac{1}{f_{n-1} f_{n+1}} = \frac{1}{f_{n-1}}\left(\frac{1}{f_{n}} - \frac{1}{f_{n+1}}\right). \] Verify by simplifying each side to see if they equal.
02

Verifying Equation (a)

Perform algebraic manipulation: the right-hand side \( \frac{1}{f_{n-1}} \left( \frac{1}{f_{n}} - \frac{1}{f_{n+1}} \right) = \frac{1}{f_{n-1} f_{n}} - \frac{1}{f_{n-1} f_{n+1}} \). Simplify to get \( \frac{f_{n+1} - f_{n}}{f_{n-1} f_{n} f_{n+1}} \). Since by definition of the Fibonacci sequence, \( f_{n+1} = f_{n} + f_{n-1} \), it simplifies to \( \frac{f_{n-1}}{f_{n-1} f_{n} f_{n+1}} = \frac{1}{f_{n} f_{n+1}} \). Thus, both sides are equal.
03

Solving Equation (b) using Equation (a)

Use equation (a) to compute the sum \( \sum_{n=2}^{x} \frac{1}{f_{n-1} f_{n+1}} \). This sum is telescoping, meaning most terms cancel each other out, simplifying to \( \frac{1}{f_{1} f_{2}} - \frac{1}{f_{x} f_{x+1}} \). Plug in \( f_1 = f_2 = 1 \) to find that \( \frac{1}{1\cdot1} - \frac{1}{f_{x} f_{x+1}} = 1 - \frac{1}{f_{x} f_{x+1}} \). As \( x \to \infty \), \( \frac{1}{f_{x} f_{x+1}} \to 0 \), confirming the equation holds true.
04

Show Equation (c) Using Equation (a)

Here, we need to show that \( \sum_{n=2}^{\infty} \frac{f_{n}}{f_{n-1} f_{n+1}} = 2 \). Recognize it as the result of summing equation (a). By rearranging \( \frac{1}{f_{n-1}} - \frac{1}{f_{n+1}} \), you get partial fraction form, where the original sum telescopes to \( \frac{1}{f_1} = 1 + \frac{1}{f_2} = 2 \), because after all cancellations, only the first and the second terms remain at infinity.

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Key Concepts

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

Telescoping Series
A telescoping series is a unique type of series where many terms cancel out, leaving only a few terms that determine the sum. These series are particularly useful in problems involving fractions or sequences. In the given exercise, the expression \( \frac{1}{f_{n-1} f_{n+1}} = \frac{1}{f_{n-1} f_{n}} - \frac{1}{f_{n} f_{n+1}} \) is a telescoping fraction. When this expression is summed over a range, most intermediate terms from the sequence will cancel out.
Here's how it works: if you expand this type of expression across a series, terms in the middle eventually cancel, simplifying the sum. This property is used to reduce complex sums in exercises, providing a simpler result.
Understanding the concept of a telescoping series can make analyzing and solving such problems much easier and quicker.
Algebraic Manipulation
Algebraic manipulation involves using algebraic techniques to simplify or rearrange expressions. In the context of the provided solution, algebraic manipulation is used to show the equivalence of the original identity in the exercise.
By rewriting \( \frac{1}{f_{n-1} f_{n+1}} \) into \( \frac{1}{f_{n-1}} \left( \frac{1}{f_{n}} - \frac{1}{f_{n+1}} \right) \), you can see how the terms are separated and set up for cancellation when summed.
This technique allows us to confirm that each side of an equation really expresses the same quantity, aiding in identifying and proving relationships like the ones found in Fibonacci sequence-related problems. It is an essential skill in mathematics that aids in solving complex equations, reducing them to simpler, more understandable forms.
Infinite Series
An infinite series is a sum of infinitely many terms. In the exercise, when dealing with a series like \( \sum_{n=2}^{\infty} \frac{1}{f_{n-1} f_{n+1}} \), this concept comes into play. Such series require careful analysis because you're adding an infinite number of terms.
This specific series simplifies due to the telescoping nature of its terms. When you start calculating the series, terms cancel sequentially, leaving a finite result even though the series itself is infinite. In this case, as shown in step 3, the leftover terms yield a result of 1.
Infinite series are foundational in calculus and analysis, providing tools to approximate functions, calculate sums, and have applications ranging from engineering to economics.
Sequence Identities
Sequence identities are mathematical expressions that hold true for specific sequences under certain conditions. These identities can simplify calculations and help prove other mathematical statements.
In the context of the Fibonacci sequence, sequence identities like \( f_{n+1} = f_{n} + f_{n-1} \) are fundamental. They describe intrinsic properties of the sequence and are used extensively in calculations and proofs.
When dealing with any sequence problem, recognizing and applying these identities can drastically simplify solving processes and provide insights into the nature of the sequence. They are powerful tools in both theoretical and applied mathematics.

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