4.052. Given two infinite geometric progressions with a common ratio $|q|<1$, differing only in the sign of their common ratios. Their sums are respectively equal to $S_{1}$ and $S_{2}$. Find the sum of the infinite geometric progression formed by the squares of the terms of any of the given progressions.
Solution.

From the condition, we have $S_{1}=\frac{b_{1}}{1-q}, S_{2}=\frac{b_{1}}{1+q},|q|<1$. Further,

$$
b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+\ldots=b_{1}^{2}+b_{1}^{2} q^{2}+b_{1}^{2} q^{4}+\ldots=b_{1}^{2}\left(1+q^{2}+q^{4}+\ldots\right)
$$

The expression in parentheses is the sum of the terms of an infinitely decreasing geometric progression, where the first term is one and the common ratio is $q^{2}$. By the formula for the sum of the terms of a geometric progression, we have

$$
b_{1}^{2}\left(1+q^{2}+q^{4}+\ldots\right)=b_{1}^{2} \cdot \frac{1}{1-q^{2}}=\frac{b}{1-q} \cdot \frac{b}{1+q}=S_{1} \cdot S_{2}
$$

Answer: $S_{1} \cdot S_{2}$.