N7 (10-6, UK) Let $[x]$ denote the greatest integer not exceeding $x$. For any positive integer $n$, compute the sum
$$
\sum_{k=0}^{\infty}\left[\frac{n+2^{k}}{2^{k+1}}\right] \text {. }
$$
Solution 1 First, for any positive integer $n$, when $2^{k}>n$, there must be $\frac{n}{2^{k+1}}<\frac{1}{2}$. Therefore, when $2^{k}>n$, $\left[\frac{n+2^{k}}{2^{k+1}}\right]=0$. This means that for any positive integer $n$, there always exists a non-negative integer $l=\left[\log _{2} n\right]$, such that:
When $k>l$, $\left[\frac{n+2^{k}}{2^{k+1}}\right]=0$;
When $k \leqslant l$, $\left[\frac{n+2^{k}}{2^{k+1}}\right] \neq 0$.
Therefore, the required sum is
$$
\sum_{k=0}^{\infty}\left[\frac{n+2^{k}}{2^{k+1}}\right]=\sum_{k=0}^{l}\left[\frac{n+2^{k}}{2^{k+1}}\right] .
$$

Let $a_{k}=\left[\frac{n+2^{k}}{2^{k+1}}\right] \quad(k=0,1, \cdots, l)$, then $a_{k}$ is the largest positive integer $m$ with the following properties:

That is
$$
\begin{array}{c}
m \leqslant \frac{n+2^{k}}{2^{k+1}}, \\
2^{k+1} m-2^{k} \leqslant n .
\end{array}
$$

Consider the sequence $\left\{2^{k}(2 m-1)\right\}$ (where $k=0,1, \cdots, l, m=1,2, \cdots, a_{k}$ ). Since each $k$ corresponds to $a_{k}$ terms in this sequence, the number of terms in this sequence is $\sum_{k=0}^{l} a_{k}$.

On the other hand, each positive integer $s$ not greater than $n$ can be uniquely expressed in the form $s=$ $2^{k}(2 m-1)$, and in this expression, $0 \leqslant k \leqslant l, 1 \leqslant m \leqslant a_{k}$. This indicates that each number in $1,2, \cdots, n$ appears in the sequence $\left\{2^{k}(2 m-1)\right\}$, and only once, so the number of terms in this sequence is $n$, thus
$$
\sum_{k=0}^{l} a_{k}=n,
$$

Therefore, the required sum is
$$
\sum_{k=0}^{\infty}\left[\frac{n+2^{k}}{2^{k+1}}\right]=n .
$$

Solution 2 Represent $n$ in binary, we have
$$
n=a_{0} \cdot 2^{r}+a_{1} \cdot 2^{r-1}+\cdots+a_{r-1} \cdot 2+a_{r},
$$

where $a_{i}=0$ or $1(i=0,1, \cdots, r)$, and $a_{0} \neq 0$. Thus
$$
\begin{array}{l}
{\left[\frac{n+2^{k}}{2^{k+1}}\right]=a_{0} \cdot 2^{r-(k+1)}+a_{1} \cdot 2^{r-(k+2)}+\cdots+a_{r-(k+1)}+a_{r-k},} \\
k=0,1,2, \cdots, r, \\
{\left[\frac{n+2^{k}}{2^{k+1}}\right]=0, \text { when } k>r \text { . }}
\end{array}
$$

Therefore, the sum
$$
\begin{aligned}
\sum_{k=0}^{\infty}\left[\frac{n+2^{k}}{2^{k+1}}\right]= & \sum_{k=0}^{r}\left[\frac{n+2^{k}}{2^{k+1}}\right] \\
= & a_{0} 2^{r-1}+a_{1} 2^{r-2}+\cdots+a_{r-1}+a_{r}+a_{0} 2^{r-2}+a_{1} 2^{r-3} \\
& +\cdots+a_{r-1}+\cdots+a_{0} \cdot 2^{0}+a_{1}+a_{0} \\
= & a_{0}\left(1+1+2+\cdots+2^{r-1}\right)+a_{1}\left(1+1+2+\cdots+2^{r-2}\right) \\
& +\cdots+a_{r-1}(1+1)+a_{r} \\
= & a_{0} 2^{r}+a_{1} 2^{r-1}+\cdots+a_{r-1} 2+a_{r} \\
= & n .
\end{aligned}
$$

Solution 3 Let $\{x\}=x-[x]$, then
$$
\left[\frac{1}{2}+\left\{\frac{x}{2}\right\}\right]=\left\{\begin{array}{ll}
0, & \text { if }\left\{\frac{x}{2}\right\}<\frac{1}{2}, \\
1, & \text { if }\left\{\frac{x}{2}\right\} \geqslant \frac{1}{2} .
\end{array}\right.
$$

Thus $\left[\frac{1}{2}+\left\{\frac{x}{2}\right\}\right]=\left[2\left\{\frac{x}{2}\right\}\right]$,
so
$$
\begin{aligned}
{\left[\frac{1}{2}+\frac{x}{2}\right] } & =\left[2\left\{\frac{x}{2}\right\}\right]+\left[\frac{x}{2}\right] \\
& =[x]-\left[\frac{x}{2}\right] .
\end{aligned}
$$

Using (1) we get
$$
\begin{aligned}
\sum_{k=0}^{\infty}\left[\frac{n+2^{k}}{2^{k+1}}\right] & =\sum_{k=0}^{\infty}\left(\left[\frac{n}{2^{k}}\right]-\left[\frac{n}{2^{k+1}}\right]\right) \\
& =\left[\frac{n}{2^{0}}\right] \\
& =n .
\end{aligned}
$$