5. (4 points) In a regular triangular pyramid (not a regular tetrahedron), the area of the base is four times smaller than the area of a lateral face. The height of the pyramid is 130 cm.

Let's construct the following (infinite) sequence of spheres. Let \( S_{1} \) be the inscribed sphere of this pyramid. Then \( S_{2} \) is the sphere that touches the lateral faces of the pyramid and the sphere \( S_{1} \); \( S_{3} \) is the sphere that touches the lateral faces of the pyramid and the sphere \( S_{2} \), not equal to \( S_{1} \), and so on; \( S_{n+1} \) is the sphere that touches the lateral faces of the pyramid and the sphere \( S_{n} \), not equal to \( S_{n-1} \).

Find the total volume of all these spheres.
Answer: $\frac{8788}{2598} \pi$ dm $^{3}$.

Solution: Let the area of the base be $S$, the height of the pyramid be $h$, and the radius of the sphere $S_{1}$ be $r$. Then, expressing the volume of the tetrahedron in two ways, we get the equality $\frac{1}{3} \cdot 13 S \cdot r=\frac{1}{3} S h$, from which $r=\frac{h}{13}=10$ cm.

Since the pyramid is regular, the centers of the spheres $S_{1}$ and $S_{2}$, as well as the point of their tangency, lie on the height of the pyramid, and the plane tangent to the spheres at the point of their tangency is parallel to the base of the pyramid. Therefore, it cuts off a similar pyramid from the original pyramid with a height of $h-2 r=110$ cm. Thus, the radius of the sphere $S_{2}$ is $\frac{11}{13}$ of the radius of the sphere $S_{1}$. Similarly, the radius of each subsequent sphere is $\frac{11}{13}$ of the radius of the previous sphere.

Thus, their volumes form an infinite decreasing geometric progression with the first term $\frac{4}{3} \pi r^{3}$ and the common ratio $\frac{11^{3}}{13^{3}}$. Its sum is $\frac{\frac{4}{\frac{4}{3} \pi 10^{3}}}{1-\frac{11^{3}}{13^{3}}}=\frac{\frac{4}{3} \pi 1000}{\frac{866}{2197}}$ cm $^{3}$ or $\frac{8788}{2558} \pi$ dm $^{3}$.