1. ## Compounded Annually

Paul plans to contribute to a retirement fund. He will invest $500 on each birthday from age 25 to 64 inclusive. That is, he will make 40 contributions to the fund. The retirement fund pays interest on the invesments at the rate of 8% per annum, compounded annually. How much money will be in Paul's fund on his 65th birthday? Thanks 2. Originally Posted by DivideBy0 Paul plans to contribute to a retirement fund. He will invest$500 on each birthday from age 25 to 64 inclusive. That is, he will make 40 contributions to the fund. The retirement fund pays interest on the invesments at the rate of 8% per annum, compounded annually. How much money will be in Paul's fund on his 65th birthday?

Thanks
i think, this is an Annuity type problem..
Amt at 65th bday $\displaystyle = \$ 500 \cdot \frac{(1+i)^n - 1}{d}$, where n is the number of years and$\displaystyle d = \frac{i}{1+i}$using this formula, i got:$\displaystyle \approx \$139,890.52$

3. Thanks kalagota, but what is $\displaystyle i$ in the formula? Also, how would you derive this formula?

Edit: nvm I get $\displaystyle i$, but why is $\displaystyle d=\frac{i}{i+1}$

4. Originally Posted by DivideBy0
Thanks kalagota, but what is $\displaystyle i$ in the formula? Also, how would you derive this formula?
oww, i is the interest in decimal form..
derive, i'll try to explain..

for a period of n years, and you invested an amount say k every year at the beginning of each year, which is compounded annually with an interest i times 100%.. suppose your last payment was at the beginning of the nth year..

say, you placed k at t=0(this is the beginning of the 1st year); after the end of nth year, it will earn a total of $\displaystyle k(1+i)^n$.. now, placing another k at t=1, after the end of nth year, that money will earn a total of $\displaystyle k(1+i)^{n-1}$.. and so on.. until the beginning of the nth year (or the end of the (n-1)th year), you will place the last k, and at the end of the nth year, it will earn $\displaystyle k(1+i)$.. thus you will have the sum:

$\displaystyle \underbrace {k(1+i)^{n} + k(1+i)^{n-1} + ... + k(1+i)}_{n \, years}$..

then using some derivations, you will get the formula..

5. Oh I understand now, thanks kalagota

I actually interpreted the problem differently

I thought he kept adding money to the total balance WHILE the total balance was being charged interest. This would mean evaluating $\displaystyle f^{40}(0)$, where $\displaystyle f(x) =1.08(x+ 500)$.
I mean it seems odd that he would have 40 different accounts with 500 dollars in them each.

Are you sure this isn't what it meant?

6. Originally Posted by DivideBy0
Oh I understand now, thanks kalagota

I actually interpreted the problem differently

I thought he kept adding money to the total balance WHILE the total balance was being charged interest. This would mean evaluating $\displaystyle f^{40}(0)$, where $\displaystyle f(x) =1.08(x+ 500)$.
I mean it seems odd that he would have 40 different accounts with 500 dollars in them each.

Are you sure this isn't what it meant?
i think no.. even my brother who is an accountant interpreted the way i did..
anyways, when i computed it, the answer was almost the same (the new answer exceed by 500).. byt my God, i have to make a program using scilab to compute it.. Ü