# Thread: Some really easy math, which I can't do

1. ## Some really easy math, which I can't do

I was given some information:

An animal population grows by 8% every year, and in addition 40 individual move there at the end of each year. At the beginning of the first year the size of the population is 2000.

Determine a recursive sequence, which gives the number of individuals in the population at the beginning of the year n.

Thus

an= 1,08*an-1+40(from the previous year)

Then the exericise asks for the general equation for the population. By deriving the equation, I get $0,8*{n-1}*2000+ 250(1-0,8^{n-1})$

Yet the answer in the book states

$1750*08^{n-1}+250$.

And I don't know how they get it.

Any help is hugely appreciated, since I have a test soon, and currently am barely reaching the level required to pass.

2. Hello,

How can you derive the equation ? oO

Let's take the problem by the beginning.

$a_1 = 2000$

At year 2, you'll have $a_2=2000*1.08+40$

At year 3, you'll have $a_3=(2000*1.08+40)*1.08 + 40=2000*1.08^2+40*(1+1.08)$

Let's continue to year 4, just to see :

$a_4=(2000*1.08^2+40*(1+1.08))*1.08+40=2000*1.08^3+ 40+40*(1.08+1.08^2)$

So it seems that the general formula will be :

$a_n=2000*1.08^{n-1}+40 \sum_{k=0}^{n-2} 1.08^k$

$a_n=2000*1.08^{n-1}+ 40 \dfrac{1-1.08^{n-1}}{1-1.08}$

$a_n=2000*1.08^{n-1}+ 40 \dfrac{1.08^{n-1}-1}{.08}$

$a_n=2000*1.08^{n-1}+500(1.08^{n-1}-1)$

$a_n=2500*1.08^{n-1}-500$

And now, i admit i don't know why i don't have the same result :'(

3. Hello, Moo!

And thank you for your reply again. Sorry for the bad translation of deriving a formula, my book is in Finnish, so I sometimes have to use inprecise defintions. I am also baffled by how the book got a different result, but if we for example calculate the number of animals at the beginning of year ten both formulas gve the same answer.

4. Hey

Never mind !

$0,8^{n-1}*2000+ 250(1-0,8^{n-1})$

This is correct (it was your formula, i just corrected what seemed to be a typo ??)

If you develop :

$2000*0.8^{n-1}+250-250*0.8^{n-1}=(2000-250)*0.8^{n-1}+250$

Which gives you the result of your book ^^

Now i just have to read again my demo

5. Originally Posted by Moo
Hey

Never mind !

$0,8^{n-1}*2000+ 250(1-0,8^{n-1})$

This is correct (it was your formula, i just corrected what seemed to be a typo ??)

If you develop :

$2000*0.8^{n-1}+250-250*0.8^{n-1}=(2000-250)*0.8^{n-1}+250$

Which gives you the result of your book ^^

Now i just have to read again my demo

Wow, thank you so much!

6. Now, i'm curious to know how you got the formula

Am a bit tired and this problem is scaring me xD

7. Well, as you can see I did many typos in my original equation, sorry

But I got it by simple simulation
a= 2000
q= 0,8
b= 40

Year Amount of animals at the beginning of the year

1 a
2 q(a+b)=qa+qb
3 q(qa+qb+b)= q^2a+q^2b+qb
4 q^3a+q^3b+q^2b+qb+b

so we have $q^{n-1}a+ \frac{b+(1-q^{n-1}}{1-q)}$

And then plug in the values.

At least that's what my book told me to do, I couldn't have figured it out by myself.