3. Mr. Alex plays the following game: At the beginning, there are nine empty boxes arranged in a row. Each operation, 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 $n$).
3.2 $\sum_{j=0}^{8} \mathrm{C}_{n}^{j}-1$.

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 initially $k$ boxes are all empty, and finally one box contains the number $2^{n}$, while the other $k-1$ boxes are empty. We call operation (1) "insertion" and operation (2) "merge".
If $k=1$, then $m(n, 1)=1$.
If $k \geqslant 2$, assume Alex performs at least one merge operation. Then, just before the last operation, exactly two boxes contain the number $2^{n-1}$, and the other $k-2$ boxes are empty. Color one of these $2^{n-1}$ blue and the other red. Trace back Alex's operations and color the numbers according to the following rules: if $a$ and $b$ are merged 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 reverse of a merge operation (since the reverse of an 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). We call an insertion of a blue number or a merge of two blue numbers a "blue operation," and similarly define a "red operation."

All sequences of blue operations can be repeated on another $k$ boxes arranged in a row, such that finally one box contains 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 finally one box contains 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, we can first perform $m(n-1, k)$ operations on $k$ empty boxes to get one box containing $2^{n-1}$, and the other $k-1$ boxes are empty. Then, perform $m(n-1, k-1)$ operations on the remaining $k-1$ empty boxes to get one box containing $2^{n-1}$, and the other $k-2$ boxes are empty. Finally, perform one merge to get one box containing $2^{n}$, and the other $k-1$ boxes are 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
$m(n, k)=m(n-1, k)+m(n-1, k-1)+1$.
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, we prove that for all non-negative integers $n$ and positive integers $k$, we have
$m(n, k)=2 \sum_{j=0}^{k-1} \mathrm{C}_{n}^{j}-1$.
We 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
$m(n+1,1)=1=2 \mathrm{C}_{n+1}^{0}-1$,
equation (2) holds.
For $k \geqslant 2$, by equation (1) and the induction hypothesis, we get
$$
\begin{array}{l}
m(n+1, k)=m(n, k)+m(n, k-1)+1 \\
=2 \sum_{j=0}^{k-1} \mathrm{C}_{n}^{j}-1+2 \sum_{j=0}^{k-2} \mathrm{C}_{n}^{j}-1+1 \\
=2 \sum_{j=1}^{k-1} \mathrm{C}_{n}^{j}+2 \sum_{j=0}^{k-2} \mathrm{C}_{n}^{j}+2 \mathrm{C}_{n}^{0}-1 \\
=2 \sum_{j=1}^{k-1} \mathrm{C}_{n}^{j}+2 \sum_{j=1}^{k-1} \mathrm{C}_{n}^{j-1}+2 \mathrm{C}_{n+1}^{0}-1 \\
=2 \sum_{j=1}^{k-1} \mathrm{C}_{n+1}^{j}+2 \mathrm{C}_{n+1}^{0}-1 \\
=2 \sum_{j=0}^{k-1} \mathrm{C}_{n+1}^{j}-1 .
\end{array}
$$