  c. The Fibonacci Series for High School StudentsA 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 haven / m  =  (n + m) / n             as m becomes very large.From our previous exercise,  a = n / m.  Thena  =  n / m   =   (n + m) / n   =   1 + m / n   =   1 + 1 / aThis 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  =  0Then 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. Main Menu