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 Explicitly Stated (Posted on 2004-08-18)
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?

 No Solution Yet Submitted by Gamer Rating: 4.0000 (5 votes)

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 Solution | Comment 2 of 11 |

I think this counts.

I started by looking at the pattern of As Bs and Cs after 5 years. Those were:
x = 1 :    A
x = 2 :    A*B + C = A*B^1 + C(1) = A*B^(x-1) + C(B^0)
x = 3 :   (A*B+C)B+C = A*B^2 + C(B+1)
= A*B^(x-1) + C(B^1 + B^0)
x = 4 :   (A*B^2+CB+C)B+C = A*B^3 + C(B^2+B+1)
= A*B^(x-1) + C(B^2 + B^1 + B^0)
x = 5 :   (A*B^3+C*B^2+CB+C)B+C
= A*B^4 + C(B^3+B^2+B+1)
= A*B^(x-1) + C(B^3 + B^2 + B^1 + B^0)

So here’s what I got as a "generality"
For year x>1,
M(x) = A*B^(x-1) + C* (the sum from j=0 to j=x-2 of B^j)

I didn’t know if there was a more straightforward equivalent of that summation. So I looked it up online (sorry!). It turns out that (the sum from j=0 to j=n of r^j) = [1-r^(n+1)]/(1-r). So in this case, r=B and n=x-2. So the sum = [1-B^(x-2+1)]/(1-B) = [1-B^(x-1)]/(1-B)

For x = 1, M(x) = A
For x > 1, M(x) = A*B^(x-1) + C*[1 – B^(x-1)]/(1–B)

 Posted by nikki on 2004-08-18 10:04:19

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