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?

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.
