3.1. Find the largest integer solution of the inequality $2^{x}+2^{x+1}+\ldots+2^{x+2000}<1$.
Answer: -2001.

Solution. By factoring out $2^{x}$ and using the formula for the sum of the first terms of a geometric progression, we can transform the inequality to $2^{x} \cdot\left(2^{2001}-1\right)<1$. For $x \geqslant-2000$, this inequality is not satisfied. If, however, $x=-2001$, we get $1-2^{-2001}<1$, which is true.