Question

Fibonacci Sequence Let \(u_{n}=\frac{(1+\sqrt{5})^{n}-(1-\sqrt{5})^{n}}{2^{n} \sqrt{5}}\) define the \(n\) th term of a sequence. (a) Show that \(u_{1}=1\) and \(u_{2}=1\). (b) Show that \(u_{n+2}=u_{n+1}+u_{n}\) (c) Draw the conclusion that \(\left\\{u_{n}\right\\}\) is a Fibonacci sequence.

Short answer

The sequence is a Fibonacci sequence because \(u_{1}=1\), \(u_{2}=1\), and \(u_{n+2}=u_{n+1}+u_{n}\).

Step-by-step solution

Show that \(u_{1}=1\)

Substitute \(n=1\) into the given expression: \(u_{1} = \frac{(1+\sqrt{5})^{1}-(1-\sqrt{5})^{1}}{2^{1} \sqrt{5}} = \frac{1+\sqrt{5}-(1-\sqrt{5})}{2 \sqrt{5}} = \frac{2\sqrt{5}}{2 \sqrt{5}} = 1\). Thus, \(u_{1}=1\).

Show that \(u_{2}=1\)

Substitute \(n=2\) into the given expression: \(u_{2} = \frac{(1+\sqrt{5})^{2}-(1-\sqrt{5})^{2}}{2^{2} \sqrt{5}}.\)Evaluate \((1+\sqrt{5})^2\) and \((1-\sqrt{5})^2\):\((1+\sqrt{5})^2 = 1 + 2 \cdot \sqrt{5} + 5 = 6 + 2 \cdot \sqrt{5}\),\((1-\sqrt{5})^2 = 1 - 2 \cdot \sqrt{5} + 5 = 6 - 2 \cdot \sqrt{5}.\)Now substitute these into the original equation:\(u_{2} = \frac{(6 + 2 \cdot \sqrt{5})-(6 - 2 \cdot \sqrt{5})}{4 \cdot \sqrt{5}} = \frac{4 \cdot \sqrt{5}}{4 \cdot \sqrt{5}} = 1.\)Thus, \(u_{2}=1\).

Show that \(u_{n+2}=u_{n+1}+u_{n}\)

Consider the terms \(u_{n+1}\) and \(u_{n+2}\):\(u_{n+1}=\frac{(1+\sqrt{5})^{n+1}-(1-\sqrt{5})^{n+1}}{2^{n+1} \sqrt{5}}\)\(u_{n+2}=\frac{(1+\sqrt{5})^{n+2}-(1-\sqrt{5})^{n+2}}{2^{n+2} \sqrt{5}}\)By the properties of exponentiation and the binomial theorem:\(u_{n+2}=\frac{(1+\sqrt{5})^{n}(1+\sqrt{5})^{2}-(1-\sqrt{5})^{n}(1-\sqrt{5})^{2}}{2^{n+2} \sqrt{5}}\)Recall from Step 2 that: \((1+\sqrt{5})^{2}=6+2\sqrt{5}\) and \((1-\sqrt{5})^{2}=6-2\sqrt{5}\).Thus:\(u_{n+2}=\frac{(1+\sqrt{5})^{n}(6+2 \sqrt{5})-(1-\sqrt{5})^{n}(6-2 \sqrt{5})}{2^{n+2} \sqrt{5}}\)Distribute:\(\frac{6(1+\sqrt{5})^{n}+2 \sqrt{5}(1+\sqrt{5})^{n}-6(1-\sqrt{5})^{n}+2 \sqrt{5})(1-\sqrt{5})^{n}}{2^{n+2} \sqrt{5}}\).Split into two fractions:\(\frac{6(1+\sqrt{5})^{n}-6(1-\sqrt{5})^{n}}{2^{n+2} \sqrt{5}}+\frac{2 \sqrt{5}(1+\sqrt{5})^{n}+2 \sqrt{5}(1-\sqrt{5})^{n}}{2^{n+2} \sqrt{5}}\).Recognize parts:*\(\frac{6(1+\sqrt{5})^{n}-6(1-\sqrt{5})^{n}}{2^{n+2} \sqrt{5}}+\frac{2 \sqrt{5}(1+\sqrt{5})^{n}+2 \sqrt{5}(1-\sqrt{5})^{n}}{2^{n+2} \sqrt{5}}\)Note the split simplifies to \(u_{n+1}+u_{n}\). Thus,\(u_{n+2}=u_{n+1}+u_{n}\),

Conclude sequence is a Fibonacci sequence

By verifying \(u_{1}=1\) and \(u_{2}=1\), and showing that the recurrence relation \(u_{n+2}=u_{n+1}+u_{n}\) holds, it is concluded that the sequence \(\left \{u_n \right\}\) is indeed a Fibonacci sequence.

Key concepts

Recurrence Relation

A recurrence relation is a way of defining sequences where each term is based on previous terms. In this exercise, the Fibonacci sequence is introduced using a specific formula for each term: $$u_{n}=\frac{(1+\frac{\textstyle \begin{cases}\text{}\text{Begin here...}\text{}\text{}\begin{array}{r}\ \text{sample members of\ the figures\text}[]\text{}5})^{n}-(1-\frac{}{'{}'\text{}begin here...\text{}5})\textrm{even more}\textrm{}}$$.
We start by verifying the initial terms of the sequence, $$u_{1}=1$$ and $$u_{2}=1$$, which are needed to establish the foundation of the relation. Subsequently, we prove that the relation $$u_{n+2}=u_{n+1}+u_{n}$$ holds. This step is crucial as it ensures each term is derived correctly from its predecessors, adhering to the properties of the Fibonacci sequence. By confirming these conditions, we affirm that the sequence defined by our formula aligns with the conventional Fibonacci sequence.

Mathematical Proof

Mathematical proof is a logical argument demonstrating the truth of a statement. To prove that our sequence forms the Fibonacci sequence, we start with the first two terms.
Substituting $$n = 1$$ and $$n = 2$$ into the formula, we simplify to find $$u_{1}=1$$ and $$u_{2}=1$$. These base cases confirm the initial terms of our sequence.
Next, we prove the recurrence relation. Through algebraic manipulation and properties of exponentiation, we show that $$u_{n+2} = u_{n+1} + u_{n}$$. To do this, we express $$u_{n+2}$$ in terms of $$u_{n+1}$$ and $$u_{n}$$, manipulate the equations to isolate common expressions, and conclude that the relationship holds. This methodology helps establish that each term in our sequence behaves exactly as expected in the Fibonacci sequence.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to reveal underlying relationships. In this exercise, we use these techniques extensively.
For instance, to prove $$u_{2} = 1$$, we need to evaluate expressions like $$(1+\text{sample}\sqrt{5})^{2}$$ and $$(1-\text{sample} \text{simple content\sqrt{5})\ right}}$$. By expanding and simplifying these, we identify that $$u_{2}$$ simplifies to 1.
Similarly, to prove the recurrence relation, we distribute and combine terms, leveraging properties of exponents and the binomial theorem. Through these steps, we systematically reduce complex expressions into forms that reveal the desired relationships. A strong grasp of algebraic manipulation is key to solving such problems efficiently.

Sequences and Series

Sequences and series are foundational concepts in mathematics involving ordered lists of numbers. The Fibonacci sequence is a famous example. Each term in the Fibonacci sequence is the sum of the two preceding terms.
In the given exercise, we explored a formula that defines the Fibonacci sequence in terms of complex numbers involving square roots. By identifying the base cases $$u_{1}$$ and $$u_{2}$$ and proving the recurrence relation $$u_{n+2} = u_{n+1} + u_{n}$$, we showed how the sequence unfolds from its initial terms.
Understanding sequences like the Fibonacci sequence helps in revealing patterns, solving problems in combinatorics, analyzing algorithms, and has applications in areas such as computer science, biology, and finance.