## Related questions with answers

Suppose that each pair of a genetically engineered species of rabbits left on an island produces two new pairs of rabbits at the age of 1 month and six new pairs of rabbits at the age of 2 months and every month afterward. None of the rabbits ever die or leave the island. a) Find a recurrence relation for the number of pairs of rabbits on the island n months after one newborn pair is left on the island. b) By solving the recurrence relation in (a) determine the number of pairs of rabbits on the island n months after one pair is left on the island.

Solution

Verified(a) $\textbf{Derivation recurrence relation}$

Let $a_n$ represent the number of pairs of rabbits on the island after $n$ months.

$\textbf{First case}$ The rabbits that were present on the island in the previous $(n-1)$th month will remain on the island and will not die. Since there are $a_{n-1}$ pairs of rabbits on the island in the previous month, these $a_{n-1}$ pairs are still on the island.

$a_{n-1}$

$\textbf{Second case}$ The pairs of rabbits that were born in the previous $(n-1)$th month each had 2 pairs of rabbits. Since there were $a_{n-2}$ pairs of rabbits 2 months ago and since there were $a_{n-1}$ pairs of rabbits 1 month ago, $a_{n-1}-a_{n-2}$ pairs of rabbits were born 2 months ago.

$2(a_{n-1}-a_{n-2})$

$\textbf{Third case}$ The pairs of rabbits that were born two months ago ($n-2$) or earlier, each had 6 pairs of rabbits. Since there were $a_{n-2}$ pairs of rabbits 2 months ago, each of these $a_{n-2}$ pairs had 6 pairs of rabbits.

$6a_{n-2}$

Add all number of pairs for each case:

$a_n=a_{n-1}+2(a_{n-1}-a_{n-2})+6a_{n-2}=3a_{n-1}+4a_{n-2}$

$\textbf{Initial conditions}$

At 0 months, 1 newborn pair of rabbits is left on the island:

$a_0=1$

At 1 month, the one-month-old pair of rabbits had 2 new pairs of rabbits and thus there are then $1+2=3$ pairs of rabbits on the island:

$a_1=3$

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