Answer
See explanation
Work Step by Step
**Solution Explanation**
We have a “Fibonacci‐type” rabbit‐counting problem with the following assumptions:
1. We start with one (male–female) pair of newborn rabbits at the beginning of month 0 (so \(r_0=1\)).
2. A pair of rabbits is *not* fertile in its first month of life, but from then on produces 4 *new* pairs at the end of every month.
3. No rabbits ever die.
Let \(r_n\) be the number of *pairs* of rabbits alive at the *end* of month \(n\). We aim to:
1. Find a recurrence relation for \(r_0, r_1, r_2, \dots\).
2. Compute a few terms \(r_0,r_1,r_2,r_3,\dots\).
3. Determine how many rabbits there are at the end of one year (i.e.\ after 12 months).
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## (a) Recurrence Relation
- **Initial Conditions:**
- At the end of month 0, \(r_0=1\) (the original newborn pair).
- At the end of month 1, that original pair is only 1 month old and not yet fertile, so no new pairs are born. Hence \(r_1=1\).
- **General Month \(n\ge2\):**
Any pair that is at least 1 month old will produce 4 new pairs at the end of the month. The number of pairs at least 1 month old at the end of month \(n\) is exactly the total number of pairs that existed two months earlier (end of month \(n-2\)).
Therefore, the number of *new* pairs born in month \(n\) is \(4 \cdot r_{n-2}\).
The total number of pairs at the end of month \(n\) is the number of pairs from the previous month \(\bigl(r_{n-1}\bigr)\) **plus** the new pairs:
\[
\boxed{r_n \;=\; r_{n-1} \;+\; 4\,r_{n-2}, \quad \text{for } n\ge2.}
\]
Together with \(r_0=1\) and \(r_1=1\), this completely defines the sequence.
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## (b) Compute \(r_0,r_1,r_2,r_3,\dots\)
Using the recurrence \(r_n = r_{n-1} + 4\,r_{n-2}\), with \(r_0=1\), \(r_1=1\):
1. \(r_2 = r_1 + 4r_0 = 1 + 4\cdot1 = 5.\)
2. \(r_3 = r_2 + 4r_1 = 5 + 4\cdot1 = 9.\)
3. \(r_4 = r_3 + 4r_2 = 9 + 4\cdot5 = 9 + 20 = 29.\)
4. \(r_5 = r_4 + 4r_3 = 29 + 4\cdot9 = 29 + 36 = 65.\)
5. \(r_6 = r_5 + 4r_4 = 65 + 4\cdot29 = 65 + 116 = 181.\)
Continuing in this manner up to \(r_{12}\) (the end of the 12th month) gives:
\[
\begin{aligned}
r_7 &= 181 + 4\cdot65 = 181 + 260 = 441,\\
r_8 &= 441 + 4\cdot181 = 441 + 724 = 1165,\\
r_9 &= 1165 + 4\cdot441 = 1165 + 1764 = 2929,\\
r_{10} &= 2929 + 4\cdot1165 = 2929 + 4660 = 7589,\\
r_{11} &= 7589 + 4\cdot2929 = 7589 + 11716 = 19305,\\
r_{12} &= 19305 + 4\cdot7589 = 19305 + 30356 = 49661.
\end{aligned}
\]
Thus at the end of month 12, we have \(r_{12} = 49661\) *pairs* of rabbits.
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## (c) How Many Rabbits at the End of the Year?
A “year” here is 12 months, so at the end of month 12 we have \(49661\) *pairs*. Since each pair consists of 2 rabbits:
\[
\text{Total rabbits} \;=\; 2 \,\times\, 49661 \;=\; 99322.
\]
Hence, there are \(\boxed{99{,}322}\) rabbits (or \(\,49{,}661\) pairs) at the end of the year.
