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Thread: compound interest - retirement planning (growth function, analysis)

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    Question compound interest - retirement planning (growth function, analysis)

    1. calculate the difference of depositing $2000 every year
    Last edited by shirel; Aug 28th 2007 at 12:13 AM.
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    Quote Originally Posted by shirel
    1. calculate the difference of depositing $2000 every year
    a) from age 20 through 29
    b) from age 30 through 65
    work with 3%, 5%, 7% and 10% interest rate (use geometric series).
    In year twenty you have,
    $\displaystyle A_{20}=2000$
    The next year you increase the amount by .03 thus,
    $\displaystyle A_{21}=2000+(.03)2000=2000(1.03)$
    The next year you increase the amount by .03 thus,
    $\displaystyle A_{22}=2000(1.03)+2000(1.03)(.03)$=$\displaystyle 2000(1.03)(1.03)=2000(1.03)^2$
    The next year you increase the amount by .03 thus,
    $\displaystyle A_{23}=2000(1.03)^2+2000(.03)=2000(1.03)^3$
    In general you have, $\displaystyle n\geq 20$
    $\displaystyle A_{n}=2000(1.03)^{n-20}$
    Thus, the amount you have at year 29 is,
    $\displaystyle A_{29}=2000(1.03)^{9}\approx 2610$
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    Quote Originally Posted by shirel
    2. verify the difference between two IRA accounts earning 10% with annual $2000 deposits when one charges a $30 annual fee and the other does not after 40 years.
    Do not really understand what you are trying to say. Anyway Using the same technique I said above, you have after $\displaystyle n$ years,
    $\displaystyle 2000(1.1)^n$ BUT each year you subtract 30 dollars thus, the actual amount after $\displaystyle n$ years is,
    $\displaystyle 2000(1.1)^n-30n$ while the other one does not thus nothing is subtracted thus, $\displaystyle 2000(1.1)^n$ There difference is,
    $\displaystyle 2000(1.1)^n-\left( 2000(1.1)^n-30n \right)$
    Open parantheses,
    $\displaystyle 2000(1.1)^n-2000(1.1)^n+30n=30n$
    After forty year you have $\displaystyle 30(40)=1200$.
    Thus, using the model which does not deduct payments you get 1200 more.
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