>>112697
1+3n
1+39=40 sections

total=average*period = (40+1)/2 * (13+1) sections * 20 students/section = 5740 retards that can't into basic reasoning.

>>112700
Where did you get 39 from?

>>112705
13*3

>>112700
It happens each semester so that's twice a year except for 2000 where only the spring semester gets 3 sections and the fall already had 1

So wouldn't it come out to be 1+3(29) having it be 88 sections in Spring 2014?

>>112708
I read long semester as an academic year. If it's every semester then it will be 1+3*(falls-1)+3*springs

1+3*13+3*14=82

(1+6*13)/2 * (13+1)*2 * 20 = 22120

>>112710
>(1+6*13)/2 * (13+1)*2 * 20 = 22120

How exactly did you do this? Doesn't make sense to me at all...

Number of semesters since Fall 2000 = (year-2001)*2 + if(season=spring then 1 else 2)

So number of semester since Fall 2000 in Spring 2014 will be

(2014-2001)*2 + 1 = 27

Number of sections = 3*(number of semesters)+1

3*27 + 1 = 82

The second part of the question we need to know the total number of sections offered from Fall 2000 to Fall 2013.

So first thing we need is the number of semesters at Fall 2013

(2013-2001)*2+2 = 26

Now what >>112710 intended to do but seriously fucked up...

The formula for summing the first n entries of a series is (n/2)(first term + nth term).

Note though that the first term starts from 0 semesters from Fall 2000. So we can't use f(x) = 3x+1

We have to adjust all our semesters up by one

semester = 1+ (year-2001)*2 + if(season=spring then 1 else 2)

and use

sections = 3(x-1)+1

So Fall 2000 turns into semester = 1 and Fall 2013 has to be called semester 27.

So that makes "n" in this case 27, not 26 (and certainly not 79). The first term of the sections series is 1. The 27th term is 79.

So, again, we're using (n/2)(first term + nth term) where n is 27.

(27/2)(1+79) = 13.5*80 = 1080

Multiply 1080 sections by 20 = 21600 students total.

You can very easily verify this by doing the math brute force in Excel or whatever other spreadsheet you want (google docs, whatever).