If a seller sold half the amount and $\frac12$ a unit to each customer, and ended up selling the entire stock. How many units were sold? The seller was asked how many cheese pieces he had sold. He replied: "Today there were $4$ buyers, each buyer bought half of the remaining cheese pieces and half of one cheese." As a result, all cheeses was sold."
How many cheeses has been sold?
All I got: $x$ - number of cheeses, $$x-\left({8 \over16}x+{1 \over64}x\right)-\left({4 \over16}x+{1 \over64}x\right)-\left({2 \over16}x+{1 \over64}x\right)-\left({1 \over16}x+{1 \over64}x\right)=0$$
But it doesn't seem right. Does anyone know how to solve this?

 A: I would work backwards. So the last buyer bought half a cheese plus half of what was left, and that was the end of the cheeses, meaning that if $x_4$ was the amount of cheese that was left right before the fourth buyer, we have:
$$x_4-\frac{x_4}{2}-\frac{1}{2}=0$$
From this, we can quickly get that $x_4 = 1$, i.e. there was $1$ cheese left before the last buyer.
OK, so moving on to the third buyer:
$$x_3-\frac{x_3}{2}-\frac{1}{2}=x_4=1$$
meaning that $x_3=3$, i.e there were $3$ cheeses left before the third buyer bought their cheese.
...can you take the last two steps?
A: Proceed backward: add half a cheese and double, four times.
$$0\to1\to3\to7\to15.$$
A: Working forward with $x$ amount of cheese: 
having served the first customer
$$\frac x2-\frac12=\frac{x-1}2$$
having served the second customer
$$\frac{x-1}4-\frac12=\frac{x-3}{4}$$
having served the third customer
$$\frac{x-3}{8}-\frac12=\frac{x-7}8$$
having served the fourth customer
$$\frac{x-7}{16}-\frac12=\frac{x-15}{16}=0.$$
From the last step:
$$x=15.$$
A: Update/edit:
There are two correct answers, 15 and 11.  Below are both laid out:
*1st buyer:  5.5 pieces + 1/2
*2nd buyer:  2.5 pieces + 1/2
*3rd buyer:  1 piece + 1/2
*4th buyer:  1/2
Total cheese sold:  11 pieces.
OR
*1st buyer:  7.5 pieces + 1/2
*2nd buyer:  3.5 pieces + 1/2
*3rd buyer:  1.5 pieces + 1/2
*4th buyer:  .5 + 1/2
Total cheese sold:  15 pieces.
Great post!
A: The iterations follow an arithmetico-geometric progression
$$x\to\\
ax+b\to\\
a(ax+b)+b=a^2x+ab+b\to\\
a(a^2x+ab+b)+b=a^3x+a^2b+ab+b\to\cdots$$
and after $n$ steps,
$$x\to a^nx+\frac{1-a^n}{1-a}b.$$
In the given case, $a=\frac12,b=-\frac12$,
$$\frac x{16}-2\left({1-\frac1{16}}\right)\frac12=0,$$ or $x=15$.
