Consider the rabbit pairs that illustrate the pattern in the Fibonacci sequence. These rabbits produce exactly 1 pair of new rabbits after reaching maturity at age 2 months. Imagine that the rabbits and all their offspring live forever. Also, imagine the field the rabbits live in can expand in size so that its side length is exactly equal to the number of pairs of rabbits living in the field. What is the side length of the field at the end of two years

46368

Explanation:

Start with two non-negative integers, f(0) and f(1). Compute f(2) = f(0) + f(1). Compute f(3) = f(1) + f(2).

Continue, creating f(n) = f(n-1) + f(n-2), where each new number is the sum of the prior two numbers in the sequence.

By convention, f(0) = 0. f(1) = 1 for our new first pair. f(2) = 1 as well, as conception just occurred. The new pair is born at the end of month 2, so during month 3, f(3) = 2. Only the initial pair produces offspring in month 3, so f(4) = 3. In month 4, the initial pair and the month 2 pair breed, so f(5) = 5. At the end of a year, Fibonacci has 144 pairs of rabbits.

1st month=1

2nd month = 1

3rd month= 1+1=2

4th month=2+1=3

5th month= 3+2=5

6th month =5+3 =8

7th month= 8+5 =13

8th month = 13+8 =21

9th month =21+13 =34

10th month= 34+21 =55

11th month =55+34=89

12th month=89+55=144

13th month =144+89=233

14th month= 233+144= 377

15th month=377+233= 610

16th month = 610+377=987

17th month =987+610 =1597

18th month =1597+987 =2584

19th Month =2584+1597= 4181

20th month =4281 + 1597= 6765

21st month =6765+ 4181= 10946

22nd month= 6765+10946= 17711

23rd month= 17711+ 10946= 28657

24th month = 28657+ 17711= 46368

Therefore side length of the field at the end of 2 years is 46368

they fuse to form they have half as many chromosomes as a somatic cell of the same individual. b

enzyme active site and substrate specificity

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