4.055. The first term of a certain infinite geometric progression with a common ratio $|q|<1$ is 1, and its sum is $S$. A new infinite geometric progression is formed from the squares of the terms of this progression. Find its sum.
## Solution.

From the condition, we have $b_{1}=1,|q|<1$. Therefore, $\frac{1}{1-q}=S, 1-q=\frac{1}{S}$, $q=1-\frac{1}{S}=\frac{S-1}{S}$.

Then $b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+\ldots=b_{1}^{2}+b_{1}^{2} q^{2}+b_{1}^{2} q^{4}+\ldots=$

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

$=\frac{S^{2}}{S^{2}-S^{2}+2 S-1}=\frac{S^{2}}{2 S-1}$.

Answer: $\frac{S^{2}}{2 S-1}$.