1. ## Population Growth

Here's the problem

The growth of a certain population is modelled by the recursion formula

$a_{n+2}=\frac{3}{2}a_{n+1}-\frac{1}{2}a_n$

and $a_1=6000$, $a_2=7000$

for n $\geq$ 1, where $a_k$ is the size of the population after period k.

I have already done the first part which is to show by induction that it is strictly increasing. The next part is

Make a conjecture for a formula for $a_n$, and prove it by induction. Hint: Consider the difference between $a_n$ and some suitable upper limit.

How do I go about that? Will the formula be something like $a_n$ = 6000+something to the power of n? Any help appreciated thank you.

2. Originally Posted by Rombie
Here's the problem

The growth of a certain population is modelled by the recursion formula

$a_{n+2}=\frac{3}{2}a_{n+1}-\frac{1}{2}a_n$

and $a_1=6000$, $a_2=7000$

for n $\geq$ 1, where $a_k$ is the size of the population after period k.

I have already done the first part which is to show by induction that it is strictly increasing. The next part is

Make a conjecture for a formula for $a_n$, and prove it by induction. Hint: Consider the difference between $a_n$ and some suitable upper limit.

How do I go about that? Will the formula be something like $a_n$ = 6000+something to the power of n? Any help appreciated thank you.
Try:

$
a_n=7000-4000 ~(1/2)^n
$

RonL

3. Thanks, 8000-4000(1/2)^n works. 8000 must be that suitable upper limit. Since it says conjecture do I need to give some reasoning for how I got there? Or is it enough to propose this formula and then prove it by induction.

How would I show that the population size is limited as n tends to infinity?

4. Originally Posted by Rombie
Thanks, 8000-4000(1/2)^n works. 8000 must be that suitable upper limit. Since it says conjecture do I need to give some reasoning for how I got there? Or is it enough to propose this formula and then prove it by induction.

How would I show that the population size is limited as n tends to infinity?
You can try some experiments to see where the sequence is going, which
should allow you to guess what the upper limit might be.

As n goes to infinity if a_n=8000-4000(1/2)^n, then a_n goes to 8000 as
(1/2)^n goes to zero. Also it approches this from below as (1/2)^n is
strictly decreasing.

RonL

5. Hello, Rombie!

I got your result . . .

The growth of a certain population is modelled by the recursion formula

$a_{n+2}\;=\;\frac{3}{2}\,a_{n+1}-\frac{1}{2}\,a_n$ . and $a_1 = 6000,\;a_2=7000$

for n $\geq$ 1, where $a_k$ is the size of the population after period $k.$

Let $a_n \:=\:X^n$

We have: . $X^{n+2} \;=\;\frac{3}{2}\,X^{n+1} - \frac{1}{2}\,X^n$

Multiply by $\frac{2}{X^n}\!:\;\;2X^2 \;=\;3X - 1$

Then: . $X^2-3X + 1\:=\:0\quad\Rightarrow\quad(X - 1)(2X - 1) \:=\:0\quad\Rightarrow\quad X \:=\:1,\:\frac{1}{2}$

Form a linear combination of these roots.
The function has the form: . $a_n \;=\;A(1^n) + B\left(\frac{1}{2}\right)^n\;=\;A + \frac{B}{2^n}$

We are told that: . $\begin{array}{ccccc}a_1 \: = \:6000\!: & A + \frac{B}{2} & = & 6000 & {\color{blue}[1]} \\
a_2 \:= \:7000\!: & A + \frac{B}{4} & = & 7000 & {\color{blue}[2]}\end{array}$

Subtract [2] from [1]: . $\frac{B}{4} \:=\:-1000\quad\Rightarrow\quad B \:=\:-4000$

Substitute into [1]: . $A -\frac{4000}{2} \:=\:6000\quad\Rightarrow\quad A \:=\:8000$

. . Therefore: . $\boxed{a_n \;=\;8000 - \frac{4000}{2^n}}$

We see that: . $\lim_{n\to\infty}a_n \;=\;\lim_{n\to\infty}\left(8000 - \frac{4000}{2^n}\right) \;=\;8000$

The population has an upper limit.

6. Thanks a lot, both of you.