Gooovanov A.C.

Find the sum

$$
\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{2^{2}}{3}\right]+\left[\frac{2^{3}}{3}\right]+\cdots+\left[\frac{2^{1000}}{3}\right]
$$
Remove the first term - it is equal to 0 - and instead of the sum of the remaining 1000 terms, consider the sum

$$
2 / 3+2^{2} / 3+2^{3} / 3+. .+2^{1000} / 3
$$

This is the sum of a geometric progression, and it equals $(2^{1001}-2)/3$. Now replace all its terms with their integer parts. Note that none of these terms are integers, but the sum of any two consecutive terms is an integer (because $2^{k} / 3 + 2^{k+1} / 3 = (3 \cdot 2^{k}) / 3 = 2^{k}$). Clearly, if the sum of two non-integer numbers is an integer, then the sum of their integer parts is less than the sum of the numbers themselves by 1 ( $[\alpha+\beta]=[[\alpha]+\{\alpha\}+[\beta]+\{\beta\}]=[\alpha]+[\beta]+[\{\alpha\}+\{\beta\}]=[\alpha]+[\beta]+[1]=[\alpha]+[\beta]+1$). Therefore, when replacing each pair of consecutive terms of our geometric progression with their integer parts, the sum decreases by 1, and since there are 1000 terms in the sum, replacing all of them with their integer parts will decrease the sum by 500.

## Answer

$(2^{1001}-2) / 3-500$.