TITLE: Can this problem be generalised mathematically?
QUESTION [11 upvotes]: I found a mathematical riddle which I solved by experiment and I would like to know is there a formula of some sort to solve these kind of problems.
You have 10 Euro. You can buy a bottle of beer for 1 Euro. You can exchange 2 empty bottles for 1 new full bottle. What is the maximum number of bottles that you can go through for your 10 Euro ?
By experiment you can go through 19 bottles for the 10 Euro.
But is there some formula to obtain 19 from the initial conditions ? I thought maybe something related to sequences or some kind of limit.
Any ideas?

REPLY [11 votes]: With $n$ € you can get $2n-1$ beers.

For $n$ € you buy and drink $n$ beers, and you have $n$ empty bottles.  
Then you take two bottles $(-2)$, exchange them for a full one, drink it and add an emptied bottle to the collection $(+1)$. You have drunk one more beer and you're left with one less bottle.  
While iterating, you loose one empty bottle per each beer drank.
Finally you stop with the last single bottle.

So you have exchanged $(n-1)$ empty bottles for $(n-1)$ beers, hence you had $n+(n-1)=2n-1$ beers and one empty bottle remains.