Problem 564

Maximal polygons

A line segment of length 2n-3 is randomly split into n segments of integer length ($n \ge 3$). In the sequence given by this split, the segments are then used as consecutive sides of a convex n-polygon, formed in such a way that its area is maximal. All of the $\binom{2n-4}{n-1}$ possibilities for splitting up the initial line segment occur with the same probability.

Let E(n) be the expected value of the area that is obtained by this procedure.
For example, for n=3 the only possible split of the line segment of length 3 results in three line segments with length 1, that form an equilateral triangle with an area of $\frac{1}{4}\sqrt{3}$. Therefore E(3)=0.433013, rounded to 6 decimal places.
For n=4 you can find 4 different possible splits, each of which is composed of three line segments with length 1 and one line segment with length 2. All of these splits lead to the same maximal quadrilateral with an area of $\frac{3}{4}\sqrt{3}$, thus E(4)=1.299038, rounded to 6 decimal places.

Let $S(k)=\displaystyle \sum_{n=3}^k E(n)$.
For example, S(3)=0.433013, S(4)=1.732051, S(5)=4.604767 and S(10)=66.955511, rounded to 6 decimal places each.

Find S(50), rounded to 6 decimal places.

最大多边形

一条长为2n-3的线段被随机分割成n条整数长度的线段（$n \ge 3$），再按照这个分割给出的顺序连接这些线段，构成面积最大的凸n边形。原线段所有共$\binom{2n-4}{n-1}$种可能的分割发生的概率相同。

记上述过程得到的多边形面积的期望值为E(n)。
例如，对于n=3，长为3的线段只能唯一地分割成3条长为1的线段，组成一个面积为$\frac{1}{4}\sqrt{3}$的等边三角形。因此E(3)=0.433013，四舍五入到6位小数。
对于n=4，共有4种不同的分割，每一种分割都由3条长为1的线段和1条长为2的线段组成。所有这些分割都能得到相同的面积为$\frac{3}{4}\sqrt{3}$的最大四边形，因此E(4)=1.299038，四舍五入到6位小数。

记$S(k)=\displaystyle \sum_{n=3}^k E(n)$。
例如，S(3)=0.433013，S(4)=1.732051，S(5)=4.604767以及S(10)=66.955511，都四舍五入到6位小数。

求S(50)，四舍五入到6位小数。