Continuous Withdrawal I'm not familiar with business math,, but right now I'm thinking of a certain scenario..let say I invested 50 000 to a bank that offers an interest of 25% ,let say monthly,, and then I eventually withdraw 2500 monthly... How will I represent the variable, let say n, that will tell me when my invested amount will tend to become 0 or eventually insufficient to withdraw,,if I continuously do the same thing over and over again ?
 I tried to represent my thought in this way...
$$n = 1$$
$$(50000)(1.25) - 2500$$
$$n = 2$$
$$((50000)(1.25) - 2500)(1.25) - 2500$$
$$n = 3$$
$$(((50000)(1.25) - 2500)(1.25) - 2500)(1.25) - 2500$$
$$.$$
$$.$$
$$.$$
$$n = n$$
$$AB^{n-1} - CB^{n-2} - CB^{n-3} - . . . - C = 0$$
where $A$ is the principal amount
$B$ is the rate
$C$ is the withdraw amount
$n$ is the number of time I withdraw
 A: You can use the following formula:
$C_0\cdot q^n-r\cdot \frac{q^n-1}{q-1}=0$
with $q=1+i$
i is the interest rate (here: per month), in your case $i=0.25$
$C_0$ is the initial capital (50,000)
r is the constant withdrawal of capital (here: per month) (2,500)
You can solve the equation above for n. The value for n has to be round down.
Remark:
Your thoughts are right. But there is a more simple way to derive the formula. You can separate the compounded initial capital and the sum of the (compounded) withdrawals.

Let S be the sum of the compounded withdrawals. It´s the value at t=n of the withdrawals. If you draw out the first withdrawal, it is not be compounded (r). If you draw out the second withdrawal, it is compounded one month (rq). The last draw out would be compounded n-1 times. 
Thus $S=\color{red}{r}+\color{blue}{rq+rq^2+\ldots +rq^{n-2} +rq^{n-1}}$
Multiplying both sides by q:
Therefore $qS=\color{blue}{rq+rq^2+\ldots +rq^{n-1}}+\color{red}{rq^n}$
Substracting the second equation by the first equation:
$S-Sq=r-rq^n$
$S(1-q)=r(1-q^n)$
$S=r\frac{1-q^n}{1-q}$
S has to have the same value like the compunded initial capital at t=n. Thus $C_0$ has to be compounded n times. This gives the equation $C_0\cdot q^n=r\cdot \frac{1-q^n}{1-q}$. This equation is similar to the equation at the top of the answer. 
A: Starting with $f\left(x\right)=Bx-C$, $x_{0}=A$ and $x_{n+1}=f\left(x_{n}\right)$ it can be shown with induction that:
$$\left(B-1\right)x_{n}=B^{n}\left[\left(B-1\right)A-C\right]+C$$
If $B>1$ and $(B-1)A-C<0$ then $x_{n+1}<x_n$ for each $n$ so that eventually $x_n<0$. 
To find the smallest integer $m$ with $x_m<0$ you should solve the inequality in $n$: $$B^{n}\left[\left(B-1\right)A-C\right]+C<0$$
Then $m$ is the smallest integer that satisfies this inequality.
Btw, in your case we have $\left(B-1\right)A-C>0$ and consequently $x_{n+1}>x_n$. 
$25$% a month?...Wow, could you please tell me where?
A: By inspection, if you earn $25$% on $50000$ monthly (ie $12500$ monthly) and you only withdraw $2500$ monthly (ie much less than the monthly interest earned), then the bank balance will never go to zero but instead keep on increasing forever!
In any case here's a general method for similar problems. 
Let
$r=$ interest rate per period (=$25$% here; period taken as month);
$P=$principal (=$50000$ here, the initial investment in the bank);
$C=$cash flow (=$2500$ here, the withdrawal per period);
$n=$number of periods until the bank balance reduces to $0$. 
This is the same as equating the present value of the cash flows over $n$ periods to the value of the principal, i.e.
$$\begin{align}
\sum_{i=1}^n\frac C{(1+r)^i}&=P\\
\frac Cr\left[1-\frac 1{(1+r)^n}\right]&=P\\
n&=-\frac{\log\left(1-\dfrac {Pr}C\right)}{\log \left(1+r\right)}\qquad\blacksquare
\end{align}$$
From the above it is required that $Pr<C$, otherwise the term in the numerator within brackets will be negative, and it will not be possible to evaluate the log of this if we consider only real numbers (this is the case using parameters in the original question).
However, if instead we have $C=15000$, then the result is $n=8.03$,i.e. the bank balance will drop to zero after the $9$th monthly withdrawal (the last withdrawal will be much less than $C$).
