Fibonacci Sequence The Fibonacci sequence is defined...

Fibonacci Sequence The Fibonacci sequence is defined recursively by $a_{n+2}=a_{n}+a_{n+1},$ where $a_{1}=1$ and $a_{2}=1$ $$ \begin{array}{l}{\text { (a) Show that } \frac{1}{a_{n+1} a_{n+3}}=\frac{1}{a_{n+1} a_{n+2}}-\frac{1}{a_{n+2} a_{n+3}}} \\ {\text { (b) Show that } \sum_{n=0}^{\infty} \frac{1}{a_{n+1} a_{n+3}}=1}\end{array} $$

Fibonacci Sequence The Fibonacci sequence is defined recursively by $a_{n+2}=a_{n}+a_{n+1},$ where $a_{1}=1$ and $a_{2}=1$
$$
\begin{array}{l}{\text { (a) Show that } \frac{1}{a_{n+1} a_{n+3}}=\frac{1}{a_{n+1} a_{n+2}}-\frac{1}{a_{n+2} a_{n+3}}} \\ {\text { (b) Show that } \sum_{n=0}^{\infty} \frac{1}{a_{n+1} a_{n+3}}=1}\end{array}
$$

Key Concepts

Fibonacci Sequence

The Fibonacci sequence is a classic example of a recursively defined sequence where each term is the sum of its two immediate predecessors. This concept serves not only as a foundation for understanding recursive structures in mathematics but also as a basis for exploring various properties of sequences, such as convergence and growth rates.

Recurrence Relations

A recurrence relation defines each term of a sequence in relation to one or more previous terms. In the Fibonacci sequence, the relation a??? = a??? + a? illustrates how the sequence builds upon itself. Understanding recurrence relations is essential for analyzing and deriving formulas for sequences, allowing for the investigation of long-term behavior and closed-form solutions.

Telescoping Series

A telescoping series is a series where consecutive terms cancel each other out in a way that simplifies the sum of the series. In problems involving the Fibonacci sequence, identifying telescoping behavior can lead to elegant proofs of certain identities and enable the straightforward computation of infinite sums, as many intermediate terms cancel, leaving only a few terms to compute.

Infinite Series Summation

Infinite series summation involves the addition of an infinite list of terms and often requires proving convergence to a finite value. In the context of Fibonacci numbers and telescoping series, demonstrating that the terms cancel effectively and that the remainder converges allows one to calculate the total sum of the series, which is a foundational technique in mathematical analysis.

Transcript

00:03 So we have a fibonacci sequence defined by a n plus 2 to be equal to a n plus 1 where we have a1 to be equal to 1 and a2 to be equal to 2 so for our first part, we'd want to show that so if i have one over a .n.

00:48 Plus 1, an plus 2 minus 1 divided by a .n.

01:01 Plus 2, a .n.

01:04 Plus 3.

01:05 And this is going to be equal to a .n.

01:11 Plus 3 minus a .n.

01:15 1 divided by a .n.

01:21 Plus 1.

01:22 A .n.

01:23 Plus 2.

01:24 You have a .n.

01:26 Plus 3.

01:28 And this is going to give me a .n.

01:32 Plus 2 divided by a .n.

01:37 Plus 1, an plus 2, an plus 2, an plus 3, which is equal to 1 over a .n plus 1 ,000, plus 3.

01:53 So we have been able to show that this.

01:57 So we started from our right side, it's equal to this.

02:04 So this here it's equal to that.

02:08 Then for our second part, which is b, we'd want to show that the summation from n equal to 0 to infinity of 1 over a .n plus 1 ,000 plus 1 ,000, it's equal to 1.

02:31 Again so this implies that let's have a sequence s n to be equal to summation k from 0 to n of 1 a k plus 1 a k plus 3 for n equal to k so this is going to be equal to my summation k equal to my summation k equal to 0 to 0 to to n, i have this.

03:11 So i have one, ak plus 1, ak plus 2, minus 1, divided by ak plus 2, ak plus 2, ak plus 3...

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