How much do we save - interest rate problem I'm trying to solve the following problem:

How much do we save during 5 years by routinely saving 50€ at the end
  of each month, with annual interest rate of 6% and a continuous
  compounding? After what time do we save 50 000€?

My calculation is $50*(1+0,06)^{60}$, but my solution is not correct. Could you help me?
Thanks
 A: The mathematical answer:
Let $a_n$ be the amount of euros you save by the end of $n$ months. Thus $a_0 = 0$, and for every $n \geq 1$, one has: $$a_n = 1.005 a_{n - 1} + 50.$$ Explanation for the recurrence relation: an annual interest rate of $6\%$ means your money increases by $0.5\%$ at the end of every month, and then you save an extra $50$ euros.
We then solve this recurrence formula: $$a_n + 10000 = 1.005 a_{n - 1} + 10050 = 1.005(a_{n - 1} + 10000).$$ Hence if we define $b_n:=a_n + 10000$, then we have $b_0 = 10000$ and $b_n = 1.005b_{n - 1}$. It is then clear that after five years (i.e. $60$ months), we will have $b_{60} = 1.005^{60} \times 10000 = 13488.5015\cdots$.
Therefore, we have $a_{60} = b_{60} - 10000 = 3488.5015\cdots$. This is the amount of money you have by the end of five years.

However!
Euros don't have a unit smaller than $0.01$ euro. Hence in reality the bank should decide how to round to integers.
Assume that the bank rounds to the nearest integer cents. Let $A_n$ be the amount of CENTS you save by the end of $n$ months. Then the formula goes like: $$A_n = round(1.005 A_{n - 1}) + 5000,$$ where $round(x)$ is the nearest integer to $x$. If you calculate $A_{60}$, you will get $348847$.
Therefore you get $3488.47$ euros by the end of five years.
YOU LOSE ABOUT $3$ CENTS!

What's worse is: the bank could (and usually will) round DOWN to the nearest integer!
In that case, you only get $3488.14$ euros by the end of five years.
YOU LOSE ABOUT $36$ CENTS!!!

EDIT: about how to solve the recurrence formula in the mathematical answer.
This is called "fixed point method". A priori, we would like to find a number $M$, so that when we define $b_n$ as $a_n + M$, we get the easier recurrence formula $b_n = 1.005 b_{n - 1}$.
Therefore we should have $a_n + M = 1.005(a_{n - 1} + M)$, or $a_n = 1.005 a_{n - 1} + 0.005M$. Compare this to our original formula, we immediately see that $0.005M$ should be equal to $50$, hence $M = 10000$.
A: Your interest is nominally $0.5\%$ per month.  When that is compounded continuously the balance is multiplied by $e^{0.005} \approx 1.00501252$  To save typing, we will call that $r$.  
There are opportunities to be off by $1$ in a problem like this.  We will start at month $0$ and assume the deposits are made at the start of the month.  We take the question to be how much money you have at the start of month $60$ before you make the deposit.
The first $50$ you deposit compounds for $60$ months, so at the end is worth $50r^{60}$.  The second $50$ you deposit compounds for $59$ months, so at the end is worth $50r^{59}$  This continues, so at the end you have $$\sum_{i=1}^{60}50r^i$$
This is a geometric series, with sum $$50\frac{r^{61}-1}{r-1}\approx 3557.34$$
To find the number of months to reach $50000$ we write
$$50000=50\frac{r^{n+1}-1}{r-1}\\
1000(r-1)=r^{n+1}-1\\
(\log(1000(r-1)+1)=(n+1)\log r\\
n+1\approx 358.77$$
So it takes about $358$ months to save $50000$
