3. Mr. Alex plays the following game: At the beginning, there are nine empty boxes arranged in a row. Each time, Alex chooses one of the following two situations:
(1) Choose any number of the form $2^{j}(j \in \mathbf{N})$ and place this number in an empty box;
(2) Choose two boxes that contain the same number (they do not need to be adjacent), let this number be $2^{j}(j \in \mathbf{N})$, then replace $2^{j}$ with $2^{j+1}$ in one of the boxes, and remove $2^{j}$ from the other box.

The game ends when one box contains the number $2^{n}(n$ is a given positive integer), and the other boxes are empty. Find the maximum number of operations Alex can perform (expressed in terms of $n$).
$$
\text { 3.2 } 2 \sum_{j=0}^{8} \mathrm{C}_{n}^{j}-1 .
$$

Consider a more general problem: replace 9 with a positive integer $k$. Let $m(n, k)$ be the maximum number of operations Alex can perform such that starting with $k$ empty boxes, one box ends up with the number $2^n$ and the other $k-1$ boxes are empty. Operation (1) is called "insertion," and operation (2) is called "merging."
$$
\text { If } k=1 \text {, then } m(n, 1)=1 \text {. }
$$

If $k \geqslant 2$, assume Alex performs at least one merge operation. Before the last operation, there are exactly two boxes with the number $2^{n-1}$, and the other $k-2$ boxes are empty. Color one of the $2^{n-1}$ blue and the other red. Trace back Alex's operations and color the numbers according to the following rules: if $a$ and $b$ merge into $c$, color $a$ and $b$ the same color as $c$. Note that in this reverse process, new numbers can only be generated by the inverse of merging (since the inverse of insertion is simply deleting the number). Therefore, all numbers that appear in the entire process are one of the two colors.

Alex's first operation is an insertion. Assume the first number inserted is blue. From this point to the last operation, at least one box contains a blue number.

Except for the last operation, there are no operations involving a blue number and a red number (since all merges are performed on numbers of the same color, and insertions involve only one number). Define an operation that inserts a blue number or merges two blue numbers as a blue operation, and similarly define a red operation.

The sequence of all blue operations can be repeated on another $k$ boxes arranged in a row, such that one box ends up with the number $2^{n-1}$ and the other $k-1$ boxes are empty. Thus, at most $m(n-1, k)$ blue operations are performed.
Next, consider the red operations.
Since each red operation involves at least one box with a blue number, the sequence of all red operations can be repeated on another $k-1$ boxes arranged in a row, such that one box ends up with the number $2^{n-1}$ and the other $k-2$ boxes are empty. Thus, at most $m(n-1, k-1)$ red operations are performed.
Therefore, $m(n, k) \leqslant m(n-1, k)+m(n-1, k-1)+1$.
Conversely, first perform $m(n-1, k)$ operations on $k$ empty boxes to get one box with $2^{n-1}$ and the other $k-1$ boxes empty. Then, perform $m(n-1, k-1)$ operations on the remaining $k-1$ empty boxes to get one box with $2^{n-1}$ and the other $k-2$ boxes empty. Finally, perform one merge to get one box with $2^n$ and the other $k-1$ boxes empty.
Therefore, $m(n, k) \geqslant m(n-1, k)+m(n-1, k-1)+1$.
Thus, for all positive integers $n$ and $k \geqslant 2$, we have
$m(n, k)=m(n-1, k)+m(n-1, k-1)+1$. (1)
When $k=1$ or $n=0$, the first operation must be an insertion of $2^n$, so
$$
m(0, k)=1, m(n, 1)=1\left(k \in \mathbf{Z}_{+}, n \in \mathbf{N}\right) .
$$

The initial values and equation (1) uniquely determine $m(n, k)$.
Finally, prove that for all non-negative integers $n$ and positive integers $k$,
$$
m(n, k)=2 \sum_{j=0}^{k-1} \mathrm{C}_{n}^{j}-1 .
$$

Use mathematical induction on $n$.
For any $m$ and positive integer $k$, by $m(0, k)=1$, equation (2) holds.

Assume for some fixed non-negative integer $n$ and all positive integers $k$, equation (2) holds.
When $k=1$, since
$$
m(n+1,1)=1=2 \mathrm{C}_{n+1}^{0}-1 \text {, }
$$

equation (2) holds.
For $k \geqslant 2$, by equation (1) and the induction hypothesis, we get
$$
\begin{array}{l}
m(n+1, k)=m(n, k)+m(n, k-1)+1 \\
=2 \sum_{j=0}^{k-1} \mathrm{C}_{n}^{j}-1+2 \sum_{j=0}^{k-2} \mathrm{C}_{n}^{j}-1+1 \\
=2 \sum_{j=1}^{k-1} \mathrm{C}_{n}^{j}+2 \sum_{j=0}^{k-2} \mathrm{C}_{n}^{j}+2 \mathrm{C}_{n}^{0}-1 \\
=2 \sum_{j=1}^{k-1} \mathrm{C}_{n}^{j}+2 \sum_{j=1}^{k-1} \mathrm{C}_{n}^{j-1}+2 \mathrm{C}_{n+1}^{0}-1 \\
=2 \sum_{j=1}^{k-1} \mathrm{C}_{n+1}^{j}+2 \mathrm{C}_{n+1}^{0}-1 \\
=2 \sum_{j=0}^{k-1} \mathrm{C}_{n+1}^{j}-1 .
\end{array}
$$