Example 4 As shown in Figure 5, connect isosceles right triangles on a square in an infinite repetitive process. The side length of the first square is 1. The sum of the area of the first square and the first isosceles right triangle is $S_{1}$, the sum of the area of the second square and the second isosceles right triangle is $S_{2}$, ...... the sum of the area of the $n$-th square and the $n$-th isosceles right triangle is $S_{n}$. Then, $S_{n}=$ $\qquad$ (explore, experiment, conjecture, express with an algebraic formula containing $n$).
Given that the sum of the area of the first square and the first isosceles right triangle is
$$
S_{1}=1+\frac{1}{4}=\frac{5}{4} \text {. }
$$

The sum of the area of the second square and the second isosceles right triangle is
$$
S_{2}=\frac{1}{4} \times 2+\frac{1}{4} \times 2 \times \frac{1}{4}=\frac{5}{4} \times \frac{1}{2} .
$$

The sum of the area of the third square and the third isosceles right triangle is
$$
S_{3}=\frac{1}{8} \times 2+\frac{1}{8} \times 2 \times \frac{1}{4}=\frac{5}{4} \times \frac{1}{2^{2}} .
$$

From this, we can infer that the sum of the area of the $n$-th square and the $n$-th isosceles right triangle is
$$
S_{n}=\frac{5}{4} \times \frac{1}{2^{n-1}} .
$$

Investigation: The sum of the areas of $n$ squares is
$$
1+\frac{1}{2}+\left(\frac{1}{2}\right)^{2}+\cdots+\left(\frac{1}{2}\right)^{n-1} ;
$$
The sum of the areas of $n$ isosceles right triangles is
$$
\frac{1}{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{3}+\cdots+\left(\frac{1}{2}\right)^{n-1} \text {. }
$$

If the triangle is a right triangle with a $30^{\circ}$ angle, then the sum of the areas of $n$ squares with side lengths equal to the opposite side of the $30^{\circ}$ angle is $1+\frac{1}{4}+\left(\frac{1}{4}\right)^{2}+\cdots+\left(\frac{1}{4}\right)^{n-1}$; the sum of the areas of $n$ squares with side lengths equal to the adjacent side of the $30^{\circ}$ angle is $1+\frac{\sqrt{3}}{4}+\left(\frac{\sqrt{3}}{4}\right)^{2}+\cdots+\left(\frac{\sqrt{3}}{4}\right)^{n-1}$.