
b. The Fibonacci Series for Middle School Students
The sequence of numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, … appears in the book The Da Vinci Code. The author notes that the ratio of consecutive terms appears to approach a constant: 34 / 21 = 1.619 ... 55 / 34 = 1.6176 ... 89 / 55 = 1.618 ... 144 / 89 = 1.6179 …
That constant is approximately 1.618. Thus, the series could be approximated by a function such as y_{right} = b(a^{n}), where a = 1.618.
The series extends to the left as 13, 8, 5, 3, 2, 1, 1, 0. Now we have to make use of some earlier learning. A negative number times a negative number yields a positive number, so we must have an expression for this side, which is
y_{left} = b(a)^{n}
We might guess whether n is positive or negative, or we could make use of the knowledge that a negative exponent implies division. Then we would have
y_{left} = b(1/a)^{n}
Let us line up the arithmetic series 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5 along with the Fibonacci series so that the zero elements are aligned:
7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7 13, 8, 5, 3, 2, 1, 1, 0, 1, 1, 2, 3, 5, 8, 13
Because a is greater than 1 (and 1/a is less than 1), as n becomes larger, (1/a)^{n} approaches zero, and as n becomes more negative, a^{n} becomes smaller. To match the negative signs, the function for the whole series becomes y = b[(a^{n})  (1/a)^{n}]. To calculate b, let us see what we get when n = 1:
y = 1 = b(1.618 + .618) = 2.236(b). Therefore, b = 1 / 2.236.
Thus, by pattern recognition we found the Fibonacci series to be described by the function y = [(1.618^{n})  (0.618)^{n}] / 2.236.
We have used the engineering approach of applying the simple basics of negative numbers and exponents as we recognized patterns.
The same problem as seen by a high school student

