Suppose that $a_1 = 1$, and for all $n \ge 2$, \[ a_n = a_{n-1} + 2a_{n-2} + 3a_{n-3} + \ldots + (n-1)a_1. \] Furthermore, suppose 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 $k$.

Let's think step by step and output the final answer within \boxed{}.
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