Problem 51 - Discrete Mathematics with Applications

Question

That \(b_{n}=F_{n} .\) It is called the Binet form of the \(n\) th Fibonacci number, after the French mathematician JacquesPhillipe-Marie Binet.) With \(\alpha\) and \(\beta\) as above, let \(u_{n}=\alpha^{n}+\beta^{n}, n \geq 1 .\) Verify each. $$u_{1}=1$$

Step-by-Step Solution

Answer

Given the equation \(u_n = \alpha^n + \beta^n\), where \(\alpha = \frac{1+\sqrt{5}}{2}\) and \(\beta = \frac{1-\sqrt{5}}{2}\), we need to verify if \(u_1 = 1\). Plugging in \(n=1\), we get \(u_1 = \alpha^1 + \beta^1 = \frac{1+\sqrt{5}}{2} + \frac{1-\sqrt{5}}{2}\). Simplifying, \(u_1 = \frac{2}{2} = 1\), and as expected, we've verified that \(u_1 = 1\).

Step 1: To begin, plug in the value of \(n=1\) into the equation given: \(u_{n} = \alpha^n + \beta^n\). #Step 2: Compute the values of α and β#

In order to calculate the values of \(\alpha\) and \(\beta\), recall the values given for them: \(\alpha = \frac{1+\sqrt{5}}{2}\) and \(\beta = \frac{1-\sqrt{5}}{2}\). #Step 3: Raise α and β to the power of n#

Step 2: Since \(n=1\) in this case, raising \(\alpha\) and \(\beta\) to the power of \(n\) simply gives the same values: \(\alpha^1 = \frac{1+\sqrt{5}}{2}\) and \(\beta^1 = \frac{1-\sqrt{5}}{2}\) #Step 4: Add α and β results#

Now, using the results from Step 3, add \(\alpha^1\) and \(\beta^1\) as required by the equation: \(u_{1} = \alpha^1 + \beta^1 = \frac{1+\sqrt{5}}{2} + \frac{1-\sqrt{5}}{2}\) #Step 5: Simplify the expression#

Step 3: Finally, simplify the expression by adding the two fractions together: \(u_{1} = \frac{(1+\sqrt{5})+(1-\sqrt{5})}{2} = \frac{2}{2}\) #Step 6: Verify the result#

Since \(u_{1} = \frac{2}{2} = 1\), we have successfully verified that \(u_{1} = 1\). This is in alignment with the problem statement, confirming the validity of the Binet form equation for the given value of \(n\).