A certain bank doesn't believe in interest and gives none the whole year. However, they do two things as a gift at the end of the year. They put money into your account such that it has 5 times as much as it did before. Then, they put 8 dollars in the account after that.

Jack gets one of these accounts at the start of year 1, and puts in 6 dollars. Assuming there are no other withdrawals or deposits into that account, figure out how much money is in that account at the beginning of year x, even if you don't know how much was in the account any of the previous years.

For example, on the beginning of year 1, he would have 6 dollars. On the beginning of year 2, he would have 38 dollars, and on the beginning of year 3 he would have 198 dollars.

What if you put in A dollars to start at the beginning of the first year, the bank put money into your account at the end of the year such that it was B times as much as before, and then put in C more dollars after that; how much money would you have at the beginning of year x, assuming everything else is normal and there are no withdrawals or deposits, even if you don't know how much was in the account any of the previous years?

Gamer referred me to this problem in the forums, so when I figured out the solution I figured I'd post it here where it belongs.  (Thanks again for the help!)

A: The amount you initially deposit in the account
B: The amount you multiply the balance by at the end of each year.
C: The amount you add to the new balance at the end of each year.
n: The number of years.

F(0) = A
F(1) = AB + c
F(2) = ABB + CB + C
F(3) = ABBB + CBB + CB + C
etc.

In other words:

F(n) = AB^n + C*[B^(n-1) + B^(n-2) ... + B^(n-n)]

That part in the brackets is just the sum of B^n for values of n from 0 to n-1.  I looked this up and found that it is simply (1-B^n)/(1-B).

So the final solution is:

F(n) = AB^n + C*(1-B^n)/(1-B)

For the original problem it gives the following:

A = 6
B = 5
C = 8

F(0) = 6
F(1) = 6*5^1 + 8 = 38
F(2) = 6*5^2 + 8*(1-5^2)/(1-5) = 6*25 + 8*6 = 198
F(3) = 6*5^3 + 8*(1-5^3)/(1-5) = 6*125 + 8*31 = 998
etc.

Posted by tomarken on 2006-04-24 11:54:26