Let $a = \sqrt[401]{4} - 1$ and for each $n \ge 2$, let $b_n = \binom{n}{1} + \binom{n}{2} a + \ldots + \binom{n}{n} a^{n-1}$. Find $b_{2006} - b_{2005}$.

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