c. The Fibonacci Series for High School Students
A high school student would use an algebraic approach to the problem. Let m be any term in the Fibonacci Series and n the next term. If we use the "Da Vinci Code" approach we have
n / m = (n + m) / n as m becomes very large.
From our previous exercise, a = n / m. Then
a = n / m = (n + m) / n = 1 + m / n = 1 + 1 / a
This becomes a2 = a + 1 as we multiply both sides of the equation by a. Rewriting, we get: a2- a - 1 = 0.
To solve this quadratic equation, substitute c + ½ for a, where the ½ is half the coefficient of a. Then we have
(c + ½)2 - (c + ½) - 1 = 0 c2 + c + ¼ - c - ½ - 1 = 0 c2 - 5/4 = 0
Then c = ±(v5) / 2 and a = ½ ±(v5) / 2, where we choose a = (1 + v5) / 2, which evaluates to 1.618... Since b = 1 / (a + 1/a), we have
1 / a = 2 / (1 + v5) = 2(v5 - 1) / [(1 + v5)(v5 - 1)] = 2(v5 - 1) / (5 - 1) = (v5 - 1) / 2 a + 1/a = (1 + v5) / 2 + (v5 - 1) / 2 = v5 b = 1 / v5
The Fibonacci function, therefore, is ([(1 + v5) / 2]n - [(1 - v5) / 2]n) / v5.
We took a second grade lesson and extended it through high school, making use of pattern recognition and the basic knowledge of negative numbers, fractions, and exponents--a concept called number structure. Most importantly, we developed it in a way that forced students to think rather than rely on rote memorization or "plugging numbers" into a formula. That is math education.
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