As shown in the figure, a sector ABD with center A and central angle $90 ^ { \circ }$ is drawn in a square ABCD with side length 5. Let $\mathrm { A } _ { 1 }$ be the point that divides segment AD in the ratio $3 : 2$, and let $\mathrm { B } _ { 1 }$ be the point where the line passing through $\mathrm { A } _ { 1 }$ and parallel to segment AB meets arc BD. A square $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } \mathrm { C } _ { 1 } \mathrm { D } _ { 1 }$ is drawn with segment $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 }$ as one side and meeting segment DC, and then a sector $\mathrm { D } _ { 1 } \mathrm {~A} _ { 1 } \mathrm { C } _ { 1 }$ with center $\mathrm { D } _ { 1 }$ and central angle $90 ^ { \circ }$ is drawn. Let $\mathrm { E } _ { 1 } , \mathrm {~F} _ { 1 }$ be the points where segment DC meets arc $\mathrm { A } _ { 1 } \mathrm { C } _ { 1 }$ and segment $\mathrm { B } _ { 1 } \mathrm { C } _ { 1 }$ respectively. The region enclosed by two segments $\mathrm { DA } _ { 1 } , \mathrm { DE } _ { 1 }$ and arc $\mathrm { A } _ { 1 } \mathrm { E } _ { 1 }$, and the region enclosed by two segments $\mathrm { E } _ { 1 } \mathrm {~F} _ { 1 } , \mathrm {~F} _ { 1 } \mathrm { C } _ { 1 }$ and arc $\mathrm { E } _ { 1 } \mathrm { C } _ { 1 }$ are shaded to obtain the figure $R _ { 1 }$.
In figure $R _ { 1 }$, a sector $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } \mathrm { D } _ { 1 }$ with center $\mathrm { A } _ { 1 }$ and central angle $90 ^ { \circ }$ is drawn in square $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 } \mathrm { C } _ { 1 } \mathrm { D } _ { 1 }$. Let $\mathrm { A } _ { 2 }$ be the point that divides segment $\mathrm { A } _ { 1 } \mathrm { D } _ { 1 }$ in the ratio $3 : 2$, and let $\mathrm { B } _ { 2 }$ be the point where the line passing through $\mathrm { A } _ { 2 }$ and parallel to segment $\mathrm { A } _ { 1 } \mathrm {~B} _ { 1 }$ meets arc $\mathrm { B } _ { 1 } \mathrm { D } _ { 1 }$. A square $\mathrm { A } _ { 2 } \mathrm {~B} _ { 2 } \mathrm { C } _ { 2 } \mathrm { D } _ { 2 }$ is drawn with segment $\mathrm { A } _ { 2 } \mathrm {~B} _ { 2 }$ as one side and meeting segment $\mathrm { D } _ { 1 } \mathrm { C } _ { 1 }$, and then a shaded figure is drawn and colored in square $\mathrm { A } _ { 2 } \mathrm {~B} _ { 2 } \mathrm { C } _ { 2 } \mathrm { D } _ { 2 }$ in the same way as obtaining figure $R _ { 1 }$ to obtain the figure $R _ { 2 }$. Continuing this process, let $S _ { n }$ denote the area of the shaded part in the $n$-th obtained figure $R _ { n }$. What is the value of $\lim _ { n \rightarrow \infty } S _ { n }$? [4 points]
(1) $\frac { 50 } { 3 } \left( 3 - \sqrt { 3 } + \frac { \pi } { 6 } \right)$
(2) $\frac { 100 } { 9 } \left( 3 - \sqrt { 3 } + \frac { \pi } { 3 } \right)$
(3) $\frac { 50 } { 3 } \left( 2 - \sqrt { 3 } + \frac { \pi } { 3 } \right)$
(4) $\frac { 100 } { 9 } \left( 3 - \sqrt { 3 } + \frac { \pi } { 6 } \right)$
(5) $\frac { 100 } { 9 } \left( 2 - \sqrt { 3 } + \frac { \pi } { 3 } \right)$