Two Egg problem confusion

So you need to solve n+(n-1)+(n-2)+...+1<=100, from where (n)(n+1)/2<=100 (this function transform is done with arithmetic series aka sum of an arithmetic sequence), now if you solve for n (wolframalpha: Reduce[Floor[n + n^2] >= 200, n] ) you get 14. Now you know that the first floor where you need to make the drop is 14th floor, next will be (14+14-1)th floor and whole sequence:

14; 27; 39; 50; 60; 69; 77; 84; 90; 95; 99; 100 

If you break the first egg, you go back to the last one and linearly check all options until you break the second egg, when you do, you got your answer. There is no magic.

http://mathworld.wolfram.com/ArithmeticSeries.html

You seem to be assuming equal-sized segments. For an optimal solution, if the first segment is of size N, then the second has to be of size N-1, and so on (because when you start testing the second segment, you've already dropped the egg once for the first segment).