There really is no single "obvious" choice here...
Possibly the sequence is periodic, with seven copies of -1 followed by six copies of 0, or perhaps seven -1s and seven 0s. Or maybe seven -1s, followed by six 0s, then five 1s, and so on, but after a certain point it would seem we have to have negative copies of a number, which is meaningless.
Or maybe it's not periodic, and every seventh value in the sequence is incremented by 1? Who knows?
I'll go ahead and assume the latter case, that the sequence is not periodic, since that's technically somewhat easier to manage. We can assign the following rule to the

-th term in the sequence:


for

.
So the generating function for this sequence might be

As to what is meant by "closed form", I'm not sure. Would this answer be acceptable? Or do you need to find a possibly more tractable form for the coefficient not in terms of the floor function?
Alright, 14 oz. cup of coffee contains 20% milk. If you add 7 ounces to that mixture, your answer would be:
14 oz = 0.2 milk
+7 oz. = 28
28/14 = 2
0.2*2 = 0.4, written as a percent will be 40%
40% milk.
Here is our profit as a function of # of posters
p(x) =-10x² + 200x - 250
Here is our price per poster, as a function of the # of posters:
pr(x) = 20 - x
Since we want to find the optimum price and # of posters, let's plug our price function into our profit function, to find the optimum x, and then use that to find the optimum price:
p(x) = -10 (20-x)² + 200 (20 - x) - 250
p(x) = -10 (400 -40x + x²) + 4000 - 200x - 250
Take a look at our profit function. It is a normal trinomial square, with a negative sign on the squared term. This means the curve is a downward facing parabola, so our profit maximum will be the top of the curve.
By taking the derivative, we can find where p'(x) = 0 (where the slope of p(x) equals 0), to see where the top of profit function is.
p(x) = -4000 +400x -10x² + 4000 -200x -250
p'(x) = 400 - 20x -200
0 = 200 - 20x
20x = 200
x = 10
p'(x) = 0 at x=10. This is the peak of our profit function. To find the price per poster, plug x=10 into our price function:
price = 20 - x
price = 10
Now plug x=10 into our original profit function in order to find our maximum profit:
<span>p(x)= -10x^2 +200x -250
p(x) = -10 (10)</span>² +200 (10) - 250
<span>p(x) = -1000 + 2000 - 250
p(x) = 750
Correct answer is C)</span>