Problem 49 - Discrete Mathematics with Applications

Question

The \(n\) th term \(b_{n}\) of a number sequence is defined by \(b_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta},\) where \(\alpha=(1+\sqrt{5}) / 2\) and \(\beta=(1-\sqrt{5}) / 2\) are solutions of the equation \(x^{2}=x+1\) Verify each. $$b_{2}=1$$

Step-by-Step Solution

Answer

The given formula for the nth term of the sequence is \(b_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}\), with constants \(\alpha = \frac{1 + \sqrt{5}}{2}\) and \(\beta = \frac{1 - \sqrt{5}}{2}\). We need to verify that \(b_2 = 1\). Substituting \(n = 2\) into the formula and the values of \(\alpha\) and \(\beta\), we have \(b_2 = \frac{\left(\frac{1 + \sqrt{5}}{2}\right)^2 - \left(\frac{1 - \sqrt{5}}{2}\right)^2}{\frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2}}\). Simplifying this expression, we obtain \(b_2 = \frac{4\sqrt{5}}{2\sqrt{5}} = 2\). However, this result contradicts the given fact that \(b_2 = 1\). Therefore, we cannot verify that \(b_2 = 1\) based on the provided information and formula.

Step 1: Write the given values

The formula for the nth term of the sequence is \(b_n = \frac{\alpha^n - \beta^n}{\alpha - \beta}\), with given constants \(\alpha = \frac{1 + \sqrt{5}}{2}\) and \(\beta = \frac{1 - \sqrt{5}}{2}\). We need to verify that \(b_2 = 1\), i.e., check if the formula gives the correct result when \(n = 2\). #Step 2: Substituting n=2#

Step 2: Substituting n=2

We substitute \(n=2\) into the formula for \(b_n\) to get \(b_2 = \frac{\alpha^2 - \beta^2}{\alpha - \beta}\). Now we need to evaluate this expression using the given values of \(\alpha\) and \(\beta\). #Step 3: Substitute the values of α and β#

Step 3: Substitute the values of α and β

Substitute the given values of \(\alpha = \frac{1 + \sqrt{5}}{2}\) and \(\beta = \frac{1 - \sqrt{5}}{2}\) into the expression for \(b_2\). This gives us: \[b_2 = \frac{\left(\frac{1 + \sqrt{5}}{2}\right)^2 - \left(\frac{1 - \sqrt{5}}{2}\right)^2}{\frac{1 + \sqrt{5}}{2} - \frac{1 - \sqrt{5}}{2}}\] #Step 4: Simplify the expression#

Step 4: Simplify the expression

Now we need to simplify the expression for \(b_2\). First, let's square both the numerators: \[\left(\frac{1 + \sqrt{5}}{2}\right)^2 = \frac{1 + 2\sqrt{5} + 5}{4}\] \[\left(\frac{1 - \sqrt{5}}{2}\right)^2 = \frac{1 - 2\sqrt{5} + 5}{4}\] Next, subtract the second numerator from the first one and simplify the denominator: \[\frac{(1 + 2\sqrt{5} + 5) - (1 - 2\sqrt{5} + 5)}{(1 + \sqrt{5}) - (1 - \sqrt{5})} = \] \[\frac{4\sqrt{5}}{2\sqrt{5}}\] #Step 5: Final simplification and verification#

Step 5: Final simplification and verification

Simplify the expression for \(b_2\) to obtain the final result: \[b_2 = \frac{4\sqrt{5}}{2\sqrt{5}} = \frac{4}{2} = 2\] However, this result contradicts the given fact that \(b_2 = 1\). Therefore, we cannot verify that \(b_2 = 1\) based on the provided information and formula.

Key Concepts

Binet's Formula

Binet's formula is a closed-form expression for finding the nth term of the Fibonacci sequence, which is named after the mathematician Jacques Philippe Marie Binet. Only instead of dealing with the Fibonacci sequence directly, our scenario involves a general sequence defined by the roots of a quadratic equation similar to the one that describes the golden ratio, which is intimately connected to Fibonacci numbers.

Using Binet's formula, we can express the nth term without the need for recursion, which involves directly calculating prior terms. By substituting specific values for \(\alpha\) and \(\beta\), which are the roots of the characteristic equation of the sequence, we can calculate any term quickly and efficiently. However, in our original exercise, there seems to be a slight mistake in the final step of verification, and this highlights the importance of careful calculation when using Binet's formula.

Recurrence Relations

Recurrence relations are equations that recursively define a sequence of values: each term of the sequence is defined as a function of the preceding terms. In the context of our exercise, the sequence \(b_n\) is defined by such a relationship where the nth term is derived from its predecessors using specific fixed values of the roots of the quadratic equation \(x^2 = x + 1\).

Understanding recurrence relations is crucial for solving many kinds of sequence problems in mathematics, as they describe the systematic, step-by-step approach to get from one term to the next. They have broad applications, especially in computer science for designing algorithms and in mathematics for solving complex problems through recursive functions.

Geometric Sequences

A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. While the sequence in our exercise is not a geometric sequence, understanding this type of sequence is instrumental for grasping the general concept of sequences and series.

Geometric sequences are characterized by the consistency of the multiplier from one term to the next, which allows for the creation of general formulas for finding specific terms or sums of terms within the sequence. Although our original problem deals with a different type of sequence, the idea of regularity and pattern across terms is a theme that carries over in much of sequence analysis.

Mathematical Induction

Mathematical induction is a proof technique used to prove statements about an infinite set of objects, typically natural numbers. It consists of two steps, the base case and the induction step. First, the statement is proved for the initial value, typically \(n=1\). Then, assuming that it holds for \(n=k\) (the induction hypothesis), we prove that it must also hold for \(n=k+1\).

Although not used directly in solving our sequence with Binet's formula, mathematical induction is important for verifying properties of sequences defined by recurrence relations. If the property is shown to hold for a base case and can be proved to hold for the next term given it holds for the previous ones, then it will hold for all terms of the sequence. This tool is particularly useful when direct calculation is impractical for large values of \(n\).