<s>[INST] 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?
 [/INST] 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.
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