Suppose that $a_1 = 1$, and that for all $n \ge 2$, $a_n = a_{n-1} + 2a_{n-2} + 3a_{n-3} + \ldots + (n-1)a_1.$ Suppose furthermore that $b_n = a_1 + a_2 + \ldots + a_n$ for all $n$. If $b_1 + b_2 + b_3 + \ldots + b_{2021} = a_k$ for some $k$, find $k$.

[i]Proposed by Andrew Wu[/i]
1. We start with the given recurrence relation for \(a_n\):
   \[
   a_n = a_{n-1} + 2a_{n-2} + 3a_{n-3} + \ldots + (n-1)a_1
   \]
   for \(n \geq 2\), and \(a_1 = 1\).

2. We need to express \(a_n\) in terms of the sum of previous terms. Notice that:
   \[
   a_n = a_{n-1} + 2a_{n-2} + 3a_{n-3} + \ldots + (n-1)a_1
   \]
   can be rewritten as:
   \[
   a_n = (a_{n-1} + a_{n-2} + \ldots + a_1) + (a_{n-2} + 2a_{n-3} + \ldots + (n-2)a_1)
   \]
   The first part of the sum is the sum of all previous \(a_i\) terms, which is \(b_{n-1}\). The second part is the same as \(a_{n-1}\) but shifted by one index. Therefore:
   \[
   a_n = b_{n-1} + a_{n-1}
   \]

3. Next, we define \(b_n\) as the sum of the first \(n\) terms of \(a\):
   \[
   b_n = a_1 + a_2 + \ldots + a_n
   \]
   From the recurrence relation, we have:
   \[
   b_n = a_{n+1} - a_n
   \]
   for \(n \geq 2\).

4. We also know that:
   \[
   b_1 = a_1 = 1
   \]
   and:
   \[
   b_2 = a_1 + a_2 = 1 + a_2
   \]

5. We need to find \(k\) such that:
   \[
   b_1 + b_2 + b_3 + \ldots + b_{2021} = a_k
   \]
   Using the relationship \(b_n = a_{n+1} - a_n\), we sum the series:
   \[
   b_1 + b_2 + b_3 + \ldots + b_{2021} = a_2 + (a_3 - a_2) + (a_4 - a_3) + \ldots + (a_{2022} - a_{2021})
   \]
   Notice that this is a telescoping series, where all intermediate terms cancel out, leaving:
   \[
   b_1 + b_2 + b_3 + \ldots + b_{2021} = a_{2022}
   \]

6. Therefore, \(k = 2022\).

The final answer is \( \boxed{2022} \)