7.1. (13 points) Find $\frac{S_{1}}{S_{2}}$, where

$$
S_{1}=\frac{1}{2^{18}}+\frac{1}{2^{17}}-\frac{1}{2^{16}}+\ldots+\frac{1}{2^{3}}+\frac{1}{2^{2}}-\frac{1}{2}, \quad S_{2}=\frac{1}{2}+\frac{1}{2^{2}}-\frac{1}{2^{3}}+\ldots+\frac{1}{2^{16}}+\frac{1}{2^{17}}-\frac{1}{2^{18}}
$$

(in both sums, the signs of the terms alternate as: ,,,,,,,,$++-++-++- \ldots$ ).
Answer: -0.2.

Solution. For the sum $S_{1}$ we have

$$
\frac{1}{2}\left(\frac{1}{2}\left(\frac{1}{2} S_{1}-\frac{1}{2^{19}}\right)-\frac{1}{2^{19}}\right)+\frac{1}{2^{19}}=S_{1}-\frac{1}{2^{3}}-\frac{1}{2^{2}}+\frac{1}{2}
$$

Solving this equation for $S_{1}$, we get

$$
S_{1}=\frac{1-2^{18}}{7 \cdot 2^{18}}
$$

Similarly, for the sum $S_{2}$ we find

$$
2\left(2\left(2 S_{2}-1\right)-1\right)+1=S_{2}-\frac{1}{2^{16}}-\frac{1}{2^{17}}+\frac{1}{2^{18}} \quad \Longrightarrow \quad S_{2}=\frac{5 \cdot\left(2^{18}-1\right)}{7 \cdot 2^{18}}
$$

Therefore, $\frac{S_{1}}{S_{2}}=\frac{-1}{5}=-0.2$.