Fibonacci posed the following problem: Suppose that rabbits live forever and that every month each pair produces a new pair which becomes productive at age 2 months. If we start with one newborn pair, how many pairs of rabbits will we have in the nth month? show that the answer is fn, where {fn} is the Fibonacci sequence defined in Example 3(c)

Let an= fn+1/fn and show that an-1=1+1/an-2.Assuming that{an} is convergent, find its limit.

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Answer:

Let f_n be the number of rabbit pairs at the beginning of each month. We start with one pair, that is f_1 = 1. After one month the rabbits still do not produce a new pair, which means f_2 = 1. After two months a new born pair appears, that is f_3 = 2, and so on. Let now n [tex]\geq[/tex] 3 be any natural number. We have that f_n is equal to the previous amount of pairs f_n-1 plus the amount of new born pairs. The last amount is f_n-2, since any two month younger pair produced its first baby pair. Finally we have  

f_1 = f_2 = 1,f_n = f_n-1 + f_n-2 for any natural n  [tex]\geq[/tex] 3.