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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?

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    I don'$t$ think the oil is added all at once--it is added once a week at the end of the week.2012-12-07

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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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    Just checked 1st ten cases--the explicit formula is correct...2012-12-07
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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 $3$.

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    wow... I would have never thought of that approach. Much better than mine (with regards to time usage).2012-12-07