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.

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 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   3.  

1. 3.4/4

2. 0.8/7

3. 0.5 x 3

4. 0.9 x 6

5. 10.8

6. 2.35

7. 0.567

8. 9.35

9. 0.02

10. 0.06

11. 0.16

12. 0.42

13. 0.63

14. 5.6

15. 8.1

16. 9.9

17. 0.28

18. 0.48

19. 0.27

20. 0.49

21. 0.45

22. 0.72

23. 0.9

24. 0.88

25. 0.75

26. 0.16

27. 0.328

28. 0.495

29. 0.72

30. 1.068