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Math Help - Some really easy math, which I can't do

  1. #1
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    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.



    Thank you in advance!
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  2. #2
    Moo
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    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 :'(
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  3. #3
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    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.
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  4. #4
    Moo
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    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
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  5. #5
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    Quote Originally Posted by Moo View Post
    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!
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  6. #6
    Moo
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    Now, i'm curious to know how you got the formula

    Am a bit tired and this problem is scaring me xD
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  7. #7
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    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.
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