Cigarretes are expensive!(compound interest problem)

1. Aug 10, 2006

Gablar16

I've been circling his problem for a little while and cannot find how to approach it.

A typical smoker spends 55 dollars a month on cigarrettes. Suppose that the smokers invest at the end of the month that same amount in a savings account at 4.8% compounded monthly. How much money will be in the account at the end of 40 years?

I translated that from spanish so it might be a little unclear, polease dont hesitate to ask for clarifycation.

I know that the compound interest formula is A= P(1+(r/n)^nt but the problem is that everytime I make a deposit I have to add last month amount with interest plus this months amount and it keeps mounting every month. I think the problem lies in P. I Tried plugging in all the numbers and if I invest only the first 55 dollars I will have 2305.60 at the end of the 40 years but I know thats wrong, as the real amount should be much higher. Any push in the right direction would me appreciated greatly. Thanks

2. Aug 10, 2006

jasc15

I just scratched this out on paper, and im trying to get a general solution for this: F = [1+(i/n)^nt]*A + [1+(i/n)^nt-1]*A + [1+(i/n)^nt-2]*A + ...
so its like calculating n different accounts, each with the same principle investment, just with one less compounding term for each successive account

3. Aug 10, 2006

Gablar16

Shouldn't A change over time too?

4. Aug 10, 2006

Ronnin

5. Aug 10, 2006

Gablar16

Thanks for the link Ronnin it really helped a lot.

I found a different formula in here
http://mathforum.org/dr.math/faq/faq.interest.html" [Broken]

but they both gave me the same result.
The website you gave me used (a - ar^(n+1))/(1-r) where;

a=deposit (55)
r=the rate (1+(.048\12))
n=479 (from the geometric sequence?)

I then substituted and

(55-55(1.004)^480)\(-.004)=79679.7

The other formula P = M([1+(i/q)]^nq-1)(q/i) where;

M= 55
i= .048
q=12
n=40

gave me the same result.

My only confusion is that I tried to verify with a web calculator http://www.dinkytown.net/java/CompoundSavings.html" [Broken], but it gave a number that it is a bit higher, can someone verify for me? Thanks again

Last edited by a moderator: May 2, 2017
Show that $\lim_{n\to\infty}(1+\frac{r}{n})^n=e^r$ Aug 18, 2017