So the base and height of S 1 are both half of the base and height of triangle ABC, so S 1=S/4.

The second space is a little more complicated, but it is actually very simple.

Triangle D 1E 1D2 is similar to triangle D2BC, and the similarity ratio is 1: 2 (d 1e 1 is half of BC).

So e1d2: e1b =1:3.

d2e 2 = 1/3×BC = 2/3×d 1e 1

The cardinal number of S2 is 2/3 of S 1, and the height of s 2 is 2/3 of S 1.

So S2=4/9×S 1.

And so on, S3=9/ 16×S2.

Did you find a pattern?

So Sn =/kloc-0 /×1/4× 4/9× 9/16× ... × n? /(n+ 1)？ = 1/(n+ 1)？

(Note: It can be divided in the middle. The key point is that the number corresponding to n and s N is n+ 1 )