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 + for a, where the is half the coefficient of a.  Then we have


             (c + )2 -  (c + )  -  1  =  0
        c
2 +  c  +    - c  -   -  1  =  0
                                      c
2 -  5/4  =  0

Then c = (v5) / 2  and a = (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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