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

1. ## compound interest - retirement planning (growth function, analysis)

1. calculate the difference of depositing $2000 every year 2. 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,
$A_{20}=2000$
The next year you increase the amount by .03 thus,
$A_{21}=2000+(.03)2000=2000(1.03)$
The next year you increase the amount by .03 thus,
$A_{22}=2000(1.03)+2000(1.03)(.03)$= $2000(1.03)(1.03)=2000(1.03)^2$
The next year you increase the amount by .03 thus,
$A_{23}=2000(1.03)^2+2000(.03)=2000(1.03)^3$
In general you have, $n\geq 20$
$A_{n}=2000(1.03)^{n-20}$
Thus, the amount you have at year 29 is,
$A_{29}=2000(1.03)^{9}\approx 2610$

3. 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 $n$ years,
$2000(1.1)^n$ BUT each year you subtract 30 dollars thus, the actual amount after $n$ years is,
$2000(1.1)^n-30n$ while the other one does not thus nothing is subtracted thus, $2000(1.1)^n$ There difference is,
$2000(1.1)^n-\left( 2000(1.1)^n-30n \right)$
Open parantheses,
$2000(1.1)^n-2000(1.1)^n+30n=30n$
After forty year you have $30(40)=1200$.
Thus, using the model which does not deduct payments you get 1200 more.