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Math Help - Geometric series and their application to a tricky question! Help!

  1. #1
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    Geometric series and their application to a tricky question! Help!

    Hi! I'm currently going through my math book in panic trying to soak up information before I go off to college. I want my brain back. So! I had trouble with this simple question.

    Each year a a sales-person is paid a bonus of 2000 which is banked into the same account which earns a fixed rate of interest of 6% p.a. with interest being paid annually. The amount at the end of year in the account is calculated as follows:

    Ao=2000
    A1=Ao *1.06+2000
    A2=A1* 1.06+2000 etc.

    My question would then be what the equation of the total amount of money the sales-man has in his bank account for the given year. The year then being the 'x' value here!

    I hope the question's clear enough, I'm a little out of my depth. I hope somebody out there can help me. Thanks for your time!
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  2. #2
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    Re: Geometric series and their application to a tricky question! Help!

    If you write out the money he has at the start, after 1 year, after 2 years, ect you'll see the geometric series pattern (and don't simplify any multiplication, that will make it harder to see).
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  3. #3
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    Re: Geometric series and their application to a tricky question! Help!

    I can write out quite simply, I feel like that's quite intuitive. But say I wanted to find the money he makes after 100 years. That would be a little too tedious without any equation to plug in the number of years! Any ideas?
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  4. #4
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    Re: Geometric series and their application to a tricky question! Help!

    Suppose a geometric series an has first term a and common ratio r. Call the sum of the first n terms 'T'
    T=a+ar+ar^2+...+ar^{n-1}

    Also you can see that
    Tr=ar+ar^2+ar^3+...+ar^{n}

    Taking the difference between them
    T-Tr=a+ar+ar^2+...+ar^{n-1}-ar-ar^2-ar^3-...-ar^{n}

    All the terms except the first and the last cancel out
    T-Tr=a-ar^n

    T(1-r)=a-ar^n

    T=\frac{a-ar^n}{1-r}
    Thanks from Paze
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  5. #5
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    Re: Geometric series and their application to a tricky question! Help!

    Hi! Thanks for your reply! I know the formula for the sum of a geometric sequence up to n, but the problem is they throw in the extra 2000 every year, which complicates the whole question tremendously to me. How can one incorporate the 2000 that is added to the bank every year into the series formula?
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  6. #6
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    Re: Geometric series and their application to a tricky question! Help!

    Quote Originally Posted by magiclink View Post
    the problem is they throw in the extra 2000 every year, which complicates the whole question tremendously to me. How can one incorporate the 2000 that is added to the bank every year into the series formula?
    According to Shakarri's advice, denote 1.06 by r and 2000 by a and write A0, A1, A2 and A3 explicitly through a and r only and without using any parentheses. Is the result a geometric progression? If so, what are its initial term and ratio?
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  7. #7
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    Re: Geometric series and their application to a tricky question! Help!

    Hello, magiclink!

    Each year a salesman is paid a bonus of $2000, which is banked into an account.
    which earns a fixed rate of interest of 6% p.a. with interest being paid annually.

    Find the formula for the total amount of money the salesman has in his account for year n.

    A_0 \:=\:2000

    A_1 \:=\:1.06(2000) + 2000

    A_2 \:=\:1.06[1.06(2000) + 2000] + 2000
    A_2 \:=\;1.06^2(2000) + 1.06(2000) + 2000

    A_3 \:=\:1.06[1.06^2(2000) + 1.06(2000) + 2000] + 2000
    A_3 \:=\:1.06^3(2000) + 1.06^2(2000) + 1.06(2000) + 2000

    Do you see the pattern?

    A_n \:=\:1.06^n(2000) \;+\; 1.06^{n-1}(2000) \;+\; 1.06^{n-2}(2000) \;+\; \cdots \;+\; 2000


    We have: . A_n \:=\: 2000 + 1.06(2000) + 1.06^2(2000) + \cdots + 1.06^n(2000)

    This is a geometric series.
    It has: first term a = 2000, common ratio r = 1.06, and n\!+\!1 terms.

    The sum is: . A_n \;=\;2000\,\frac{1.06^{n+1} -1}{0.06}
    Thanks from Paze
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  8. #8
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    Re: Geometric series and their application to a tricky question! Help!

    WHY YOU ARE DOING A SIMPLE COMPOUND INTEREST FORMULA SO COMPLICATE...
    JUST APPLY THE WELL KNOWN FORMULA FOR COMPOUND INTEREST....
    Pn = P0 (1+0.6)^n......
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  9. #9
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    Re: Geometric series and their application to a tricky question! Help!

    Ahh! Finally. I've understood it. I always hung up on thinking that the 2000 that's added means you'd have to change the geometric sum formula somehow. I get it now. It turns out it's still a geometric series and it's just simply about using the formula..

    Thanks everyone, much appreciated. My thick skull's been penetrated. I feel silly now!
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  10. #10
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    Re: Geometric series and their application to a tricky question! Help!

    Quote Originally Posted by MINOANMAN View Post
    WHY YOU ARE DOING A SIMPLE COMPOUND INTEREST FORMULA SO COMPLICATE...
    JUST APPLY THE WELL KNOWN FORMULA FOR COMPOUND INTEREST....
    Pn = P0 (1+0.6)^n......
    That wouldn't include adding 2000 monthly. Please don't type in caps lock.
    Thanks from Paze
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