3.21. Find all solutions of the system of equations

$$
x\left(1-\frac{1}{2^{n}}\right)+y\left(1-\frac{1}{2^{n+1}}\right)+z\left(1-\frac{1}{2^{n+2}}\right)=0
$$

where $n=1,2,3,4, \ldots$
3.21. Answer: $y=-3 x, z=2 x(x-$ is an arbitrary number).

Let's prove that the given infinite system of equations is equivalent to the system of two equations:

$$
\begin{gathered}
x+y+z=0 \\
4 x+2 y+z=0
\end{gathered}
$$

First, if we subtract the second equation, multiplied by $\frac{1}{2^{n+2}}$, from the first equation, we obtain the $n$-th equation of the original system. Second, already from the first two equations of the original system

$$
\begin{aligned}
& x\left(1-\frac{1}{2}\right)+y\left(1-\frac{1}{4}\right)+z\left(1-\frac{1}{8}\right)=0 \\
& x\left(1-\frac{1}{4}\right)+y\left(1-\frac{1}{8}\right)+z\left(1-\frac{1}{16}\right)=0
\end{aligned}
$$

the indicated two equations follow. Indeed, subtracting the second equation from the first, we get $x / 4+y / 8+z / 16=0$. Adding this equation to the second equation, we obtain $x+y+z=0$.