Let $[x]$ denote the greatest integer that is at most as large as the real number $x$. Calculate the value of the sum (1) for all positive integers $n$, and prove the correctness of the obtained result.

$$
\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{2^{2}}\right]+\ldots+\left[\frac{n+2^{k}}{2^{k+1}}\right]+\ldots
$$
I. solution. The expression to be examined contains only a finite number of non-zero terms for every prescribed $n$. Indeed, the expression in the $[ ]$ brackets of the $k+1$-th term can be written in the form

$$
\frac{1}{2}+\frac{n}{2^{k+1}}
$$

and this is a positive proper fraction for every $k$ such that

$$
\frac{n}{2^{k+1}} < 1
$$

thus its integer part is 0. (If the condition holds for some integer $k$, then it holds for all subsequent integer values, since the $2^{x}$ function is monotonically increasing.)

Let us denote the sum by $S_{n}$ and consider some values of $n$.

$$
\begin{array}{ll}
S_{1}=\left[\frac{2}{2}\right]+\left[\frac{3}{4}\right] & =1+0=1 \\
S_{2}=\left[\frac{3}{2}\right]+\left[\frac{4}{4}\right]+\left[\frac{6}{8}\right] & =1+1+0=2 \\
S_{3}=\left[\frac{4}{2}\right]+\left[\frac{5}{4}\right]+\left[\frac{7}{8}\right] & =2+1+0=3 \\
S_{4}=\left[\frac{5}{2}\right]+\left[\frac{6}{4}\right]+\left[\frac{8}{8}\right]+\left[\frac{12}{16}\right] & =2+1+1+0=4 \\
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\end{array}
$$

In these examples, increasing $n$ by 1 results in only one term of $S_{n}$ increasing by 1, while the others remain unchanged. We will show that this is always the case, thus proving the $S_{n}=n$ relationship by complete induction. Indeed,

$$
\left[\frac{n+2^{k}}{2^{k+1}}\right]=\left[\frac{n+1+2^{k}}{2^{k+1}}\right]
$$

if $n+2^{k}$ and $n+1+2^{k}$ fall between the same two multiples of $2^{k+1}$, including the lower but not the upper boundary. However, if $n+1+2^{k}$ reaches the next multiple of $2^{k+1}$, i.e., for some $m$,

$$
n+1+2^{k}=(m+1) 2^{k+1}, \quad \text { i.e., } \quad n+1=(2 m+1) 2^{k}
$$

then

$$
\left[\frac{n+2^{k}}{2^{k+1}}\right]=\left[\frac{(m+1) 2^{k+1}-1}{2^{k+1}}\right]=m \quad \text { and } \quad\left[\frac{n+1+2^{k}}{2^{k+1}}\right]=\left[\frac{(m+1) 2^{k+1}}{2^{k+1}}\right]=m+1
$$

There is only one such $k$ for each $n$: this is the exponent to which 2 must be raised to be a divisor of $n+1$, but higher powers are not. (If $n+1$ is odd, then $k=0$.)

This proves that

$$
S_{n+1}-S_{n}=1, \quad \text { so } \quad S_{n}=S_{1}+n-1=n
$$

Remark. Since the $\left[\left(n+2^{k}\right) / 2^{k+1}\right]$ term increases by 1 for those $n$ that are odd multiples of $2^{k}$ and is 1 when $n=1 \cdot 2^{k}$, this term gives the number of numbers not greater than $n$ that are divisible by $2^{k}$ but not by a higher power of 2. This is not difficult to prove directly, which gives another version of the above solution.

II. solution. We will prove the following identity below:

$$
[2 x]=[x]+\left[x+\frac{1}{2}\right]
$$

Applying this repeatedly, $n$ can be transformed as follows:

$$
\begin{gathered}
n=[n]=\left[2 \cdot \frac{n}{2}\right]=\left[\frac{n}{2}\right]+\left[\frac{n}{2}+\frac{1}{2}\right]= \\
=\left[\frac{n+1}{2}\right]+\left[\frac{n}{4}\right]+\left[\frac{n}{4}+\frac{1}{2}\right]=\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+\left[\frac{n}{8}\right]+\left[\frac{n}{8}+\frac{1}{2}\right]
\end{gathered}
$$

and generally

$$
n=\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+\ldots+\left[\frac{n+2^{l}}{2^{l+1}}\right]+\left[\frac{n}{2^{l+1}}\right]
$$

We have already seen this for $l=0,1,2$ and if it is true for some $l_{0}$ value, then applying (2) to the last term

$$
\begin{aligned}
n & =\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+\ldots+\left[\frac{n+2^{l_{0}}}{2^{l_{0}+1}}\right]+\left[\frac{n}{2^{l_{0}+1}}\right]= \\
& =\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+\ldots+\left[\frac{n+2^{l_{0}}}{2^{l_{0}+1}}\right]+\left[\frac{n}{2^{l_{0}+2}}\right]+\left[\frac{n}{2^{l_{0}+2}}+\frac{1}{2}\right]= \\
& =\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+\ldots+\left[\frac{n+2^{l_{0}}}{2^{l_{0}+1}}\right]+\left[\frac{n+2^{l_{0}+1}}{2^{l_{0}+2}}\right]+\left[\frac{n}{2^{l_{0}+2}}\right]
\end{aligned}
$$