Example 1. Let $n$ be an integer, calculate the following expression:
$$
\begin{array}{l}
{\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{2^{2}}\right]+\left[\frac{n+2^{2}}{2^{3}}\right]} \\
+\cdots .
\end{array}
$$

where the symbol $[x]$ denotes the greatest integer not exceeding $x$.
(IMO-10)
Analysis: In the expression of the problem, the denominators are powers of 2. Since $2^{\mathrm{k}}=(100 \cdots 0)_{2}$, in the binary number system, dividing by 2 is simply shifting the decimal point to the left by $k$ positions. Also, because the binary representation of a decimal integer is necessarily a binary integer, and the binary representation of a decimal pure fraction is necessarily a binary pure fraction, and vice versa, the operation of "taking the integer part" of a positive number in binary is the same as in decimal, which is to simply remove the fractional part. Thus, using binary number operations to solve this problem is very concise.
Let $n=\left(a_{m} a_{n-1} \cdots a_{1} a_{0}\right)_{2} ， a_{m}=1$. $a_{1}=0$ or $1(1=0,1,2, \cdots, m-1)$. F is, for $k=1,2, \cdots, m$ we have
$$
\begin{aligned}
& \frac{n+2^{\mathrm{k}-1}}{2^{\mathrm{k}}} \\
= & \frac{\left(a_{\mathrm{m}} a_{\mathrm{m}-1} \cdots a_{1} a_{0}\right)_{2}+\left(100^{\mathrm{k}-1 \uparrow}\right)_{2}}{(100 \cdots 0)_{2}} \\
= & \left(a_{\mathrm{m}} a_{\mathrm{m}-1} \cdots a_{\mathrm{k}} \cdot\left(a_{\mathrm{k}-1}+1\right) \cdots a_{1} a_{0}\right)_{2} .
\end{aligned}
$$

Thus, $\left[\frac{n+2^{\mathrm{k}-1}}{2^{\mathrm{k}}}\right]$
$$
=\left\{\begin{array}{ll}
\left(a_{\mathrm{m}} a_{\mathrm{m}-1} \cdots a_{\mathrm{k}}\right)_{2}+1, & a_{\mathrm{k}-1}=1, \\
\left(a_{\mathrm{m}} a_{\mathrm{m}-1} \cdots a_{\mathrm{k}}\right)_{2}, & a_{\mathrm{k}-1}=0 .
\end{array}\right.
$$

This is $\left[\frac{n+2^{\mathbf{k}-1}}{2^{\mathbf{k}}}\right]=\left(a_{m} a_{m-1} \cdots a_{k}\right)_{2}$
$$
+a_{k-1}, k=1,2, \cdots, m .
$$

Also, $\left[\frac{n+2^{\mathrm{m}}}{2^{\mathrm{m}+1}}\right]=1=a_{\mathrm{m}}, k>m+1$ when,
$\left[\frac{n+2^{\mathbf{k}-1}}{2^{k}}\right]=0$. Therefore,
$$
\begin{array}{l}
\sum_{k=1}^{\infty}\left[\frac{n+2^{k-1}}{2^{k}}\right]=\sum_{k=1}^{\mathrm{m}+1}\left[\frac{n+2^{k-1}}{2^{k}-}\right] \\
=\sum_{k=1}^{m}\left[\frac{n+2^{k-1}}{2^{k}}\right]+a_{m} \text {, } \\
=\left(a_{\mathrm{m}} a_{\mathrm{m}-1} \cdots a_{1}\right)^{2}+a_{0}+\left(a_{\mathrm{m}} a_{\mathrm{m}-1} \cdots a_{2}\right)_{2} \\
+a_{1}+\cdots+\left(a_{\mathrm{m}} a_{\mathrm{m}-1}\right)_{2}+a_{\mathrm{m}-2}+\left(a_{\mathrm{m}}\right)_{2} \\
+a_{\mathrm{m}-1}+a_{\mathrm{m}} \\
=a_{m}\left[(11 \cdots 1)_{m}+1\right]+a_{m-1}\left[(11 \cdots 1)_{2}+1\right] \\
+\cdots+a_{2}\left[(11)_{2}+1\right]+a_{1}\left[(1)_{2}+1\right]+a_{0} \\
=a_{m}(100 \cdots 0)_{m}+a_{m-1}(100 \cdots 0)_{2}+\cdots \\
+a_{2}(100)_{2}+a_{1}(10)_{2}+a_{0} \\
=\left(a_{\mathrm{m}} a_{\mathrm{m}-1} \cdots a_{2} a_{1} a_{0}\right)_{2}=n_{\text {。 }} \\
\end{array}
$$

Using the idea of binary notation, we can prove the following conclusion:
Let $n=\left(a_{\mathrm{m}} a_{\mathrm{m}-1} \cdots a_{1} a_{0}\right)_{2}, a_{\mathrm{m}}=1$. Define
$$
s(n)=a_{\mathrm{m}}+a_{\mathrm{m}-1}+\cdots+a_{1}+a_{0},
$$

then the power of 2 in $n!$ is $n-s(n)$.