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

$$
\begin{aligned}
& a_{1}+a_{1} q+a_{1} q^{2}+\ldots=S_{1}=\frac{a_{1}}{1-q} \\
& a_{1}-a_{1} q+a_{1} q^{2}-\ldots=S_{2}=\frac{a_{1}}{1+q} \\
& a_{1}^{2}+a_{1}^{2} q^{2}+a_{1}^{2} q^{4}+\ldots=S=\frac{a_{1}^{2}}{1-q^{2}}
\end{aligned}
$$

From this, it follows that

$$
S=\frac{a_{1}^{2}}{1-q^{2}}=\frac{a_{1}}{1-q} \cdot \frac{a_{1}}{1+q}=S_{1} S_{2}
$$

Answer: \( S_{1} S_{2} \).