A student wrote down the following sequence of numbers : the first number is 1, the second number is 2, and after that, each number is obtained by adding together all the previous numbers. Determine the 12th number in the sequence.
1. **Identify the sequence pattern:**
   The sequence starts with \( a_1 = 1 \) and \( a_2 = 2 \). Each subsequent number is obtained by adding all the previous numbers. Let's write down the first few terms to identify the pattern:
   \[
   \begin{align*}
   a_1 &= 1, \\
   a_2 &= 2, \\
   a_3 &= a_1 + a_2 = 1 + 2 = 3, \\
   a_4 &= a_1 + a_2 + a_3 = 1 + 2 + 3 = 6, \\
   a_5 &= a_1 + a_2 + a_3 + a_4 = 1 + 2 + 3 + 6 = 12, \\
   a_6 &= a_1 + a_2 + a_3 + a_4 + a_5 = 1 + 2 + 3 + 6 + 12 = 24.
   \end{align*}
   \]
   We observe that each term \( a_n \) for \( n \geq 3 \) can be written as \( a_n = 3 \cdot 2^{n-3} \).

2. **Prove the general term by induction:**
   We will use mathematical induction to prove that \( a_n = 3 \cdot 2^{n-3} \) for \( n \geq 3 \).

   **Base case:**
   For \( n = 3 \):
   \[
   a_3 = 3 = 3 \cdot 2^{3-3} = 3 \cdot 2^0 = 3.
   \]
   The base case holds.

   **Inductive step:**
   Assume that \( a_k = 3 \cdot 2^{k-3} \) holds for some \( k \geq 3 \). We need to show that \( a_{k+1} = 3 \cdot 2^{(k+1)-3} \).

   By the definition of the sequence:
   \[
   a_{k+1} = a_1 + a_2 + \cdots + a_k.
   \]
   Using the inductive hypothesis:
   \[
   a_{k+1} = 1 + 2 + \sum_{i=3}^{k} 3 \cdot 2^{i-3}.
   \]
   The sum of the geometric series is:
   \[
   \sum_{i=3}^{k} 3 \cdot 2^{i-3} = 3 \sum_{i=0}^{k-3} 2^i = 3 \left( \frac{2^{k-2} - 1}{2 - 1} \right) = 3 (2^{k-2} - 1).
   \]
   Therefore:
   \[
   a_{k+1} = 1 + 2 + 3 (2^{k-2} - 1) = 3 + 3 \cdot 2^{k-2} - 3 = 3 \cdot 2^{k-2}.
   \]
   Simplifying:
   \[
   a_{k+1} = 3 \cdot 2^{(k+1)-3}.
   \]
   Thus, the inductive step holds, and by induction, \( a_n = 3 \cdot 2^{n-3} \) for \( n \geq 3 \).

3. **Calculate the 12th term:**
   Using the formula \( a_n = 3 \cdot 2^{n-3} \):
   \[
   a_{12} = 3 \cdot 2^{12-3} = 3 \cdot 2^9 = 3 \cdot 512 = 1536.
   \]

The final answer is \( \boxed{1536} \).