# 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.