1. calculate the difference of depositing $2000 every year

- Apr 30th 2006, 04:01 AMshirelcompound interest - retirement planning (growth function, analysis)
1. calculate the difference of depositing $2000 every year

- Apr 30th 2006, 09:52 AMThePerfectHackerQuote:

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,

$\displaystyle A_{29}=2000(1.03)^{9}\approx 2610$ - Apr 30th 2006, 09:58 AMThePerfectHackerQuote:

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.