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?

(In reply to Solution Try by Rajal)

Whoops! That's what happens when you go too fast. The incremental amount is ((b-1)*a+c)*(b^(n-1)) the n'th time, but to combine the series of increments needs the identity:

b^0 + b^1 + b^2 + ... + b^(n-1) = (1-b^n)/(1-b)  -- (prove this to yourself by multiplying both sides by (1-b) and all the intermidiate pieces cancle) -- of course, b can't be 1 or there's a divide by 0  but that's an easy case anyway.

To use this, define d=(b-1)*a+c.

f(1) = b*a+c = a + (b-1)*a+c = a + d =

f(2)=b*(b*a+c)+c  -- blow out to:

     = b*b*a + b*c + c  --  add 0 to the front:

     = (b*a - b*a) + b*b*a + b*c + c  --  rearrange:

     = b*a+c + b*b*a - b*a + b*c  --  pull an 'a' out of the b*a:

     = a + (b-1)*a+c + b*b*a - b*a + b*c  --  factor the b & b-1

     = a + (b-1)*a+c + b*((b-1)*a+c)  -- substitute the d's

     = a + d + b*d

f(3)= a + d + b*d + b^2*d --  similar to above derivation

this continues to be generalized as f(n) = a + b^0*d + b^1*d + . . . + b^(n-1)*d

So, the answer is f(n) = a + (1-b^n)/(1-b) * ((b-1)*a+c)

(the mathamatical derivation above is harder than playing with the numbers and finding the series, but I have a hard time explaining the "Playing with Numbers" approach)

Posted by Rajal on 2004-08-19 15:16:06