4. If $[x]$ denotes the greatest integer not exceeding $x$, then $\left[\frac{1}{3}\right]+\left[\frac{2}{3}\right]+\left[\frac{2^{2}}{3}\right]+\cdots+\left[\frac{2^{1000}}{3}\right]=$
4. $\frac{1}{3}\left(2^{1001}-2\right)-500$ Detailed explanation: First, prove that "if the sum of two non-integers is an integer, then the sum of their integer parts is one less than their own sum." Also, $\frac{1}{3}+\frac{2}{3}+\frac{2^{2}}{3}+\cdots+\frac{2^{1000}}{3}=\frac{1}{3}\left(2^{1001}-2\right)$, so the original expression $=\frac{1}{3}\left(2^{1001}-2\right)-500$.