1. calculate the difference of depositing $2000 every year
In year twenty you have,Originally Posted by shirel
$\displaystyle A_{20}=2000$
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,
$\displaystyle A_{29}=2000(1.03)^{9}\approx 2610$
Do not really understand what you are trying to say. Anyway Using the same technique I said above, you have after $\displaystyle n$ years,Originally Posted by shirel
$\displaystyle 2000(1.1)^n$ 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.