What is the value of the sum $\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\left[\log _{2} 3\right]+\cdots+\left[\log _{2} 2002\right]$?
Solution. From the definition of the logarithm, it follows that

$$
\log _{2} 1=0, \quad \log _{2} 2=1, \quad \log _{2} 4=2
$$

and so on, the function always takes the next integer value at the next power of 2. We also know that the $\log _{2} x$ function is monotonically increasing. The $\left[\log _{2} x\right]$ function takes the smaller value between two powers of 2 as many times as there are integers between the two powers of 2; for example, the value 2 is taken at $\log _{2} 4, \log _{2} 5, \log _{2} 6, \log _{2} 7$.

The summation should be continued until we reach the value 2002. Since $2^{10}=1024<2002$, the last value will be 10, which the function will take at $2003-1024=979$ points.

Thus, the value of the sought sum is:

$$
\begin{aligned}
& (0+2 \cdot 1+4 \cdot 2+8 \cdot 3+\ldots+512 \cdot 9)+979 \cdot 10= \\
= & \left(\sum_{k=1}^{9} k \cdot 2^{k}\right)+9790=\left(\sum_{k=1}^{9} 2^{k}\right)+\left(\sum_{k=2}^{9} 2^{k}\right)+\ldots+\left(\sum_{k=9}^{9} 2^{k}\right)+9790= \\
= & \left(2^{10}-2\right)+\left(2^{10}-2^{2}\right)+\ldots+\left(2^{10}-2^{9}\right)+9790= \\
= & 9 \cdot 2^{10}-\left(2^{10}-2\right)+9790=2^{13}+9792=17984 .
\end{aligned}
$$