Difference Equations Tom’s old car has a major oil leak, losing 25% of the oil in the engine every week. Tom adds a quart of oil weekly. The capacity of the engine is 6 quarts of oil. In the long run, what will the oil level (in quarts) be at the end of every week before each quart of oil is added?
 A: Let's write out a few test cases and see if we can find a pattern:
   At the end of week:    Capacity (quarts):    After filling (quarts):
          1                   4.5                     5.5
          2                   4.125                   5.125
          3                   3.84375                 4.84375

Well, that wasn't too useful for pattern finding, but we can at least use it to check our work...
Let $a_i$ represent the amount of oil in the tank at then end of a given week $i$  (before filling the tank):
$$a_n = (a_{n-1}+1)\cdot\frac{3}{4}$$
for $n >= 0$ where $a_0 = 5$
Expanding:
$$a_{n+2} = (((a_{n}+1)\cdot\frac{3}{4})+1)\cdot\frac{3}{4}$$
$$a_{n+2} = (((a_{n}+1)\cdot\frac{3^2}{4^2})+\frac{3}{4})$$
$$a_{n+2} = (((\frac{3^2}{4^2}\cdot a_{n}+\frac{3^2}{4^2}))+\frac{3}{4})$$
I will generalize this to:
$$a_{n} = \left(\frac{3}{4}\right)^n\cdot a_{0}+\sum_{k=1}^{n}\frac{3^k}{4^k}$$
By geometric series:
$$a_{n} = \left(\frac{3}{4}\right)^n\cdot a_{0}+\left(\frac{3}{4}\cdot\frac{1-\left(\frac{3}{4}\right)^n}{1-\left(\frac{3}{4}\right)}\right)$$
$$a_{n} = \left(\frac{3}{4}\right)^n\cdot a_{0}+\left(3\frac{2^{2n}-3^n}{2^{2n}}\right)$$
Computing $a_3$ yields $3.84375$, thus I assume the formula is correct.
EDIT: I now see you want what happens in the "long term."  Taking the limit of the series as $n\to\infty$ evaluates to 3 quarts.
A: Let's first find the answer, on the assumption there is an answer. Let $a_n$ be the amount at the end of the $n$-th week, before adding the quart.
Then, as at the beginning of anorton's answer, we have 
$$a_{n+1}=\dfrac{3}{4}\left(1+a_n\right)\tag{$1$}$$
Assume that the limit of $a_n$ as $n\to\infty$ exists. Let that limit be $a$. Then
$$a=\frac{3}{4}(1+a).$$
Solve for $a$: we get $a=3$.
We still owe a debt, to show existence. Since we "know" that the answer is $3$, it is natural to let $a_n=b_n+3$. Substituting in $(1)$, we obtain
$$b_{n+1}=\frac{3}{4}b_n.$$
This finishes things, the "error" $b_n$ gets multiplied by $\dfrac{3}{4}$ each time, so $b_n$ has limit $0$. It follows that $a_n$ has limit $3$.   
