Example 2 Find $\sum_{0}^{1000}\left[\frac{1}{3} \times 2^{i}\right]$. (2000, Russian Mathematical Olympiad)
Analysis: According to Proposition 1, here $p=0$, the key is to find a natural number $k$, such that $\frac{1}{3} \in\left[\frac{k}{2^{1000}}, \frac{k+1}{2^{1000}}\right)$,
$$
\because \frac{1}{3}=\frac{\frac{1}{3} \times 2^{1000}}{2^{1000}}, k \leqslant \frac{1}{3} \times 2^{1000}<k+1 \text {, }
$$
$\therefore$ take $k=\frac{1}{3}\left(2^{1000}-1\right)$
$$
\begin{array}{l}
= 2^{999}-2^{998}+2^{997}-2^{996}+\cdots+ \\
2^{3}-2^{2}+2-1 \\
=2^{998}+2^{996}+\cdots+2^{2}+1
\end{array}
$$

to satisfy the condition.
For easier calculation of the result, the table is as follows:
\begin{tabular}{|c|c|c|}
\hline \begin{tabular}{l}
$\left[\frac{k}{2^{i}}\right]$
\end{tabular} & Expanded Form & \begin{tabular}{l}
$\left[\frac{k}{2^{i}}\right]+\left[\frac{k}{2^{i-1}}\right]$
\end{tabular} \\
\hline \begin{tabular}{l}
$\left[\frac{k}{2^{0}}\right]=$
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
$2^{998}+2^{996}+2^{994}+\cdots+$ \\
$2^{4}+2^{2}+1$
\end{tabular}
\end{tabular} & \multirow[b]{2}{*}{$=2^{999}-1$} \\
\hline \begin{tabular}{l}
$\left[\frac{k}{2^{1}}\right]=$
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
$2^{997}+2^{995}+2^{993}+\cdots+$ \\
$2^{3}+2^{1}$
\end{tabular}
\end{tabular} & \\
\hline \begin{tabular}{l}
$\left[\frac{k}{2^{2}}\right]=$
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
$2^{996}+2^{994}+2^{992}+\cdots+$ \\
$2^{4}+2^{2}+1$
\end{tabular}
\end{tabular} & \multirow{2}{*}{$=2^{997}-1$} \\
\hline \begin{tabular}{l}
$\left[\frac{k}{2^{3}}\right]=$
\end{tabular} & \begin{tabular}{l}
$\left\{\begin{array}{l}2^{995}+2^{993}+2^{991}+\cdots+ \\
2^{3}+2^{1}\end{array}\right.$
\end{tabular} & \\
\hline$\cdots$ & $\ldots$ & $\cdots$ \\
\hline \begin{tabular}{l}
$\left[\frac{k}{2^{996}}\right]=$
\end{tabular} & $2^{2}+1$ & \multirow{2}{*}{$=2^{3}-1$} \\
\hline \begin{tabular}{l}
$\left[\frac{k}{2^{997}}\right]=$
\end{tabular} & $2^{1}$ & \\
\hline \begin{tabular}{l}
$\left[\frac{k}{2^{998}}\right]=$
\end{tabular} & 1 & $=2-1$ \\
\hline \begin{tabular}{l}
$\sum_{0}^{1000}\left[\frac{k}{2^{i}}\right]$
\end{tabular} & & \begin{tabular}{l}
$=\frac{2^{1001}-2}{3}-500$
\end{tabular} \\
\hline
\end{tabular}

Thus, $f\left(\frac{1}{3}\right)=\sum_{0}^{1000}\left[\frac{1}{3} \times 2^{i}\right]$
$$
\begin{array}{l}
=\sum_{0}^{1000}\left[\frac{\frac{1}{3}\left(2^{1000}-1\right)}{2^{i}}\right] \\
=\frac{1}{3}\left(2^{1001}-2\right)-500 .
\end{array}
$$