Page 449, Exercise 7

a. There is little to prove here because the formula is simply
the generating function for Fibonacci numbers.

b. Observe that:

        N0 = F2 = 1
        N1 = F3 = N

      
From the text, we have 

  A. Ni = N0 + ∑Nk +1 where 0 ≤ k ≤ i-2 for all i ≥ 1 
 

  B. Fh = ∑Fk + 1, 0 ≤ k ≤ h-2 for all i > 1  

    
 Examining equation A, we know that N0 = F2.  As is also obvious from A and B, the two equations are identical except for the N0 term in  A, and the stipulation of i ≥ 1.  This means that if we simply shift the Ni formula by two, we'll get the Fibonnaci formula.  Therefore, Ni = Fi+2.

c. Φ, or the golden ratio, fits prominently in several of the fomulas for closed forms of the Fibonnaci generating function.  Φ, and its mirror, (1-√5)/2), can be used to generate the form: Fn = 1√5 (Φn- (1-√5)/2)n).  When n is even, Fn will be slightly
larger than  Φn/√5.