4A. Find the number of all subsets of three different elements from the set $\left\{2,2^{2}, 2^{3}, \ldots, 2^{2000}\right\}$ such that these elements are consecutive terms of an increasing geometric progression.
Solution. Let the three numbers from the set be $2^{a}, 2^{b}, 2^{c}, 1 \leq a < b < c \leq 2000$. For these numbers to form an increasing geometric progression, the ratio between consecutive terms must be the same, which means $2^{b-a} = 2^{c-b}$. This implies $b-a = c-b$, or $2b = a + c$. Since $a < b$, then $b-a > 0$, and the above conditions are equivalent to $b-a \leq 2000-b$, which simplifies to $a \geq 2b - 2000$. Therefore, $2b \leq 2000$, so $b \leq 1000$. Thus, for any $a, b$ such that $1 \leq a < b \leq 1000$, we can find $c$ such that $1 \leq a < b < c \leq 2000$ and $2^{a}, 2^{b}, 2^{c}$ form an increasing geometric progression. For each choice of $b$, where $2 \leq b \leq 1000$, there are $b-1$ choices for $a$, so the total number of choices for $a$ and $b$ is

$$
\sum_{b=2}^{1000}(b-1)=\sum_{b=1}^{999}(b-1)=\frac{999 \cdot 1000}{2}=499500
$$

Thus, the number of such three-element subsets is 499500.