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 $(\genfrac{}{}{0ex}{}{2n-4}{n-1})$ 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.