10.1. Summation
$$
\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{2^{2}}{3}\right]+\left[\frac{2^{3}}{3}\right]+\cdots+\left[\frac{2^{1000}}{3}\right]
$$
10. 1. $\frac{1}{3}\left(2^{1001}-2\right)-500$.

The first term is obviously 0, so we can discard it. For the remaining 1000 terms, let's first examine
$$
\frac{2}{3}+\frac{2^{2}}{3}+\frac{2^{3}}{3}+\cdots+\frac{2^{1000}}{3}
$$

This is the sum of a geometric series, whose value is $\frac{1}{3}\left(2^{1001}-2\right)$. Now let's examine the sum of their integer parts. In the above sum, no term is an integer, but the sum of any two adjacent terms is an integer (because $\frac{2^{k}}{3}+ \left.\frac{2^{k+1}}{3}=\frac{3 \cdot 2^{k}}{3}=2^{k}\right)$. It is easy to see that if the sum of two non-integers is an integer, then the sum of their integer parts is 1 less than the sum of the numbers themselves ($\alpha$
$$
\begin{array}{l}
+\beta=[\alpha]+\{\alpha\}+[\beta]+\{\beta\}=[\alpha]+[\beta]+\{a\}+ \\
\{\beta\}=[\alpha]+[\beta]+1) .
\end{array}
$$

Therefore, once we replace each pair of adjacent terms in the geometric series with their integer parts, their sum decreases by 1. Since there are 1000 terms in the sum, the sum of their integer parts is 500 less than the original sum.