Fibonacci Variation: A single pair of rabbits (male and female) is born at the b
ID: 1943270 • Letter: F
Question
Fibonacci Variation: A single pair of rabbits (male and female) is born at the beginning of a year. Asume the following conditions (which are more realistic than Fibonacci's):1) Rabbit pairs are not fertile during their first month of life but thereafter give birth to four new male/female pairs at the end of every month,
2) No rabbits die
a) Let rn=the number of pairs of rabbits alive at the end of month n, for each integer n>=1, and let r0=1. Find the recurrence relation for r0,r1,.....
b) how many rabbits will there be at the end of the year.
Explanation / Answer
The crucial observation is that the number of rabbit pairs born at the end of month k is the same as the number of pairs alive at the end of month k-2. Why? Because it is exactly the rabbit pairs that were alive at the end of month k-2 that were fertile during month k. The rabbits born at the end of month k-1 were not. so at month k-2, each pair alive at month k-1 nothing at month k-2 gives birth to a pair here F_0 = the intial number of rabbit paris = 1 and F_1 = 1 also, becuase the first apir of rabbits is not fertile until the 2nd month. Hence the complete specification of the Fibbonacci sequence for all integers k >= 2, (1) F_k = F_k-1 + F_k-2 (2) F_0 = 1, F_1 = 1 Okay now back to my problem, The number of rabbits alive at the end of the first month is still 1, so S_0 = 1; Now it won't be another 2 months after the first month until they are fertile, so at the end of month 1 they still will only have 1 pair, so S_1 = 1, and also at month 2, they still havn't mated, so S_2 = 1, but now they can mate at month 2, so at month 3 (S_3) this is where the babies start popping out. So if its 3 new male/females, I would just take 3*S_1 right? S_3 = S_0 + 3*S_1 S_4 = S_1 + 3*S_2 S_5 = S_2 + 3*S_3 ... ... S_k = S_k-3 + 3*S_k-2
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