7. (IMO-10 Problem) Let $[x]$ denote the greatest integer not exceeding $x$. Find the value of $\sum_{k=0}^{\infty}\left[\frac{n+2^{k}}{2^{k+1}}\right]$, where $n$ is any natural number.
7. For any real number, in different number systems, the integer parts are integer parts, and the fractional parts are fractional parts. Analyzing the structure of the addend is easier in binary.
Let \( n = (a_m a_{m-1} \cdots a_1 a_0)_2 \), where \( a_m = 1 \), \( a_i = 0 \) or \( 1 \), \( i = 0, 1, \cdots, m-1 \), then \(\left[\frac{n}{2^{k+1}} + \frac{1}{2}\right]\).
When \( a_k = 0 \), it is \((a_m a_{m-1} \cdots a_{k+1})\), and when \( a_k = 1 \), it is \((a_m a_{m-1} \cdots a_{k+1} + a_k)_2\), which can be reduced to the latter. Therefore,
\[
\begin{aligned}
\sum_{k=0}^{\infty} \left[\frac{n + 2^k}{2^{k+1}}\right] = & \sum_{k=0}^{m} (a_m \cdots a_{k+1})_2 + \sum_{k=0}^{m+1} (a_k)_2 = (a_m a_{m-1} \cdots a_1 + a_m a_{m-1} \cdots a_2 + \cdots + a_m a_{m-1} + \\
& a_m)_2 + (a_0 + a_1 + \cdots + a_{m-1} + a_m)_2 \\
= & a_0 + a_1 (1+1)_2 + a_2 (1+1+10)_2 + \cdots + a_m (1+1+10+\cdots+1 \underbrace{00 \cdots 0}_{\text{mik}})_2 \\
= & (a_m a_{m-1} \cdots a_1 a_0)_2 = n \text{ as required. }
\end{aligned}
\]